{
  "index": "1993-B-5",
  "type": "GEO",
  "tag": [
    "GEO",
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Show there do not exist four points in the Euclidean plane such that the\npairwise distances between the points are all odd integers.",
  "solution": "Solution 1. For real numbers \\( x \\) and \\( y \\), and for a positive integer \\( n \\), let \" \\( x \\equiv y \\) \\( (\\bmod n) \\) \" mean that \\( x-y \\) is an integer divisible by \\( n \\). Choose a coordinate system in which the four points are \\( (0,0),(a, 0),(r, s),(x, y) \\). Here \\( a \\) is an odd integer, and we may assume \\( a>0 \\). The square of an odd integer is congruent to 1 modulo 8 , so if all the pairwise distances are odd integers, we have\n\\[\n\\begin{array}{l}\nr^{2}+s^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(r-a)^{2}+s^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\nx^{2}+y^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(x-a)^{2}+y^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(x-r)^{2}+(y-s)^{2} \\equiv 1 \\quad(\\bmod 8) .\n\\end{array}\n\\]\n\nSubtracting the first two yields \\( 2 a r \\equiv a^{2}(\\bmod 8) \\). Thus \\( r \\) is a rational number whose denominator is a multiple of 2 and a divisor of \\( 2 a \\). The same is true of \\( x \\). Therefore we can multiply all coordinates by the odd integer \\( a \\), to reduce to the case where the denominators of \\( r \\) and \\( x \\) are both 2 . Then the congruence \\( 2 a r \\equiv a^{2}(\\bmod 8) \\) between integers implies \\( r \\equiv a / 2(\\bmod 4) \\). If \\( r=a / 2+4 b \\), then\n\\[\nr^{2}=a^{2} / 4+4 a b+16 b^{2} \\equiv a^{2} / 4 \\quad(\\bmod 4),\n\\]\nso the first congruence implies\n\\[\ns^{2} \\equiv 1-r^{2} \\equiv 1-a^{2} / 4 \\quad(\\bmod 4) .\n\\]\n\nSimilarly \\( x \\equiv a / 2(\\bmod 4) \\) and \\( y^{2} \\equiv 1-a^{2} / 4(\\bmod 4) \\). Also\n\\[\nx-r \\equiv a / 2-a / 2 \\equiv 0 \\quad(\\bmod 4),\n\\]\nso \\( (x-r)^{2} \\in 16 \\mathbb{Z} \\), and the last of the five congruences yields\n\\[\n(y-s)^{2} \\equiv 1-(x-r)^{2} \\equiv 1 \\quad(\\bmod 8)\n\\]\n\nWe will derive a contradiction from the congruences \\( s^{2} \\equiv y^{2} \\equiv 1-a^{2} / 4(\\bmod 4) \\) and \\( (y-s)^{2} \\equiv 1(\\bmod 8) \\) obtained above. First,\n\\[\n\\begin{aligned}\n(y+s)^{2} & \\equiv 2 y^{2}+2 s^{2}-(y-s)^{2} \\\\\n& \\equiv 2\\left(1-a^{2} / 4\\right)+2\\left(1-a^{2} / 4\\right)-1 \\\\\n& \\equiv 3-a^{2} \\\\\n& \\equiv 2 \\quad(\\bmod 4) .\n\\end{aligned}\n\\]\n\nMultiplying this integer congruence by \\( (y-s)^{2} \\equiv 1(\\bmod 8) \\) yields \\( \\left(y^{2}-s^{2}\\right)^{2} \\equiv 2 \\) \\( (\\bmod 4) \\). But \\( y^{2} \\) and \\( s^{2} \\) are rational by the beginning of this paragraph, so \\( y^{2}-s^{2} \\) is a rational number with square congruent to 2 modulo 4 . This is impossible.\n\nSolution 2.\nLemma. If \\( r=\\cos \\alpha, s=\\cos \\beta \\), and \\( t=\\cos (\\alpha+\\beta) \\), then \\( 1-r^{2}-s^{2}-t^{2}+2 r s t=0 \\). Proof. We have\n\\[\n\\begin{aligned}\n\\cos (\\alpha+\\beta) & =\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta \\\\\n(\\cos (\\alpha+\\beta)-\\cos \\alpha \\cos \\beta)^{2} & =\\sin ^{2} \\alpha \\sin ^{2} \\beta \\\\\n(t-r s)^{2} & =\\left(1-r^{2}\\right)\\left(1-s^{2}\\right)\n\\end{aligned}\n\\]\nfrom which the result follows.\nSuppose that \\( O, A, B, C \\) are four points in the plane such that \\( a=O A, b=O B \\), \\( c=O C, x=B C, y=C A, z=A B \\) are all odd integers. Using the Law of Cosines, let\n\\[\n\\begin{array}{l}\nr=\\cos \\angle A O B=\\frac{a^{2}+b^{2}-z^{2}}{2 a b} \\\\\ns=\\cos \\angle B O C=\\frac{b^{2}+c^{2}-x^{2}}{2 b c} \\\\\nt=\\cos \\angle A O C=\\frac{c^{2}+a^{2}-y^{2}}{2 c a} .\n\\end{array}\n\\]\n\nBut \\( \\angle A O B+\\angle B O C=\\angle A O C \\) as directed angles, so the lemma implies \\( 1-r^{2}-s^{2}- \\) \\( t^{2}+2 r s t=0 \\). Substituting the values of \\( r, s, t \\), and multiplying by \\( 4 a^{2} b^{2} c^{2} \\) yields\n\\[\n\\begin{aligned}\n4 a^{2} b^{2} c^{2}-c^{2}\\left(a^{2}+b^{2}-z^{2}\\right)^{2} & -a^{2}\\left(b^{2}+c^{2}-x^{2}\\right)^{2}-b^{2}\\left(c^{2}+a^{2}-y^{2}\\right)^{2} \\\\\n& +\\left(a^{2}+b^{2}-z^{2}\\right)\\left(b^{2}+c^{2}-x^{2}\\right)\\left(c^{2}+a^{2}-y^{2}\\right)=0 .\n\\end{aligned}\n\\]\n\nThe square of an odd integer is 1 modulo 4 , so we obtain\n\\[\n4-1-1-1+1 \\equiv 0 \\quad(\\bmod 4),\n\\]\na contradiction.\nSolution 3 (Manjul Bhargava). Suppose \\( \\vec{v}_{1}, \\vec{v}_{2}, \\vec{v}_{3} \\) are vectors in 3 -space. Then the volume \\( V \\) of the parallelepiped spanned by the vectors is given by \\( \\left|\\vec{v}_{1} \\cdot\\left(\\vec{v}_{2} \\times \\vec{v}_{3}\\right)\\right|= \\) \\( |\\operatorname{det} M| \\), where \\( M=\\left(\\vec{v}_{1} \\vec{v}_{2} \\vec{v}_{3}\\right) \\) is the \\( 3 \\times 3 \\) matrix with entries given by the vectors \\( \\vec{v}_{1} \\), \\( \\vec{v}_{2}, \\vec{v}_{3} \\). Since \\( \\operatorname{det} M=\\operatorname{det} M^{T} \\),\n\\[\nV^{2}=M \\cdot M^{T}=\\operatorname{det}\\left(\\begin{array}{ccc}\n\\vec{v}_{1} \\cdot \\vec{v}_{1} & \\vec{v}_{1} \\cdot \\vec{v}_{2} & \\vec{v}_{1} \\cdot \\vec{v}_{3} \\\\\n\\vec{v}_{2} \\cdot \\vec{v}_{1} & \\vec{v}_{2} \\cdot \\vec{v}_{2} & \\vec{v}_{2} \\cdot \\vec{v}_{3} \\\\\n\\vec{v}_{3} \\cdot \\vec{v}_{1} & \\vec{v}_{3} \\cdot \\vec{v}_{2} & \\vec{v}_{3} \\cdot \\vec{v}_{3}\n\\end{array}\\right)\n\\]\n\nThe volume of the tetrahedron spanned by the vectors is \\( V / 6 \\). If the edges of the tetrahedron are \\( a=\\left|\\vec{v}_{1}\\right|, b=\\left|\\vec{v}_{2}\\right|, c=\\left|\\vec{v}_{3}\\right|, x=\\left|\\vec{v}_{2}-\\vec{v}_{3}\\right|, y=\\left|\\vec{v}_{3}-\\vec{v}_{1}\\right|, z=\\left|\\vec{v}_{1}-\\vec{v}_{2}\\right| \\), then by the Law of Cosines, written as\n\\[\n\\begin{array}{c}\n2 \\vec{v}_{i} \\cdot \\vec{v}_{j}=\\left|\\vec{v}_{i}\\right|^{2}+\\left|\\vec{v}_{j}\\right|^{2}-\\left|\\vec{v}_{i}-\\vec{v}_{j}\\right|^{2}, \\\\\n8 V^{2}=\\operatorname{det}\\left(\\begin{array}{ccc}\n2 a^{2} & a^{2}+b^{2}-z^{2} & a^{2}+c^{2}-y^{2} \\\\\na^{2}+b^{2}-z^{2} & 2 b^{2} & b^{2}+c^{2}-x^{2} \\\\\na^{2}+c^{2}-y^{2} & b^{2}+c^{2}-x^{2} & 2 c^{2}\n\\end{array}\\right) .\n\\end{array}\n\\]\n\nSuppose now that there were 4 points as described in the problem. Locate one point at the origin, and let \\( \\vec{v}_{1}, \\vec{v}_{2}, \\vec{v}_{3} \\) be the vectors from the origin to the other three; they are coplanar, so \\( V=0 \\). Since squares of odd integers are congruent to 1 modulo 8 ,\n\\[\n8 V^{2} \\equiv \\operatorname{det}\\left(\\begin{array}{lll}\n2 & 1 & 1 \\\\\n1 & 2 & 1 \\\\\n1 & 1 & 2\n\\end{array}\\right) \\equiv 4 \\quad(\\bmod 8)\n\\]\nwhich gives a contradiction.\nSolution 4. Suppose as before there were 4 such points; define \\( \\vec{v}_{1}, \\vec{v}_{2}, \\vec{v}_{3} \\) as in the previous solution. Then \\( \\vec{v}_{i} \\cdot \\vec{v}_{i} \\equiv 1(\\bmod 8) \\), and from \\( (2), 2 \\vec{v}_{i} \\cdot \\vec{v}_{j} \\equiv 1(\\bmod 8) \\) for \\( i \\neq j \\) as well.\nNo three of the points can be collinear. Hence \\( \\vec{v}_{3}=x \\vec{v}_{1}+y \\vec{v}_{2} \\) for some scalars \\( x \\) and \\( y \\). Then\n\\[\n\\begin{array}{l}\n2 \\vec{v}_{1} \\cdot \\vec{v}_{3}=2 x \\vec{v}_{1} \\cdot \\vec{v}_{1}+2 y \\vec{v}_{1} \\cdot \\vec{v}_{2} \\\\\n2 \\vec{v}_{2} \\cdot \\vec{v}_{3}=2 x \\vec{v}_{2} \\cdot \\vec{v}_{1}+2 y \\vec{v}_{2} \\cdot \\vec{v}_{2} \\\\\n2 \\vec{v}_{3} \\cdot \\vec{v}_{3}=2 x \\vec{v}_{3} \\cdot \\vec{v}_{1}+2 y \\vec{v}_{3} \\cdot \\vec{v}_{2} .\n\\end{array}\n\\]\n\nSince \\( \\vec{v}_{1} \\) is not a scalar multiple of \\( \\vec{v}_{2} \\),\n\\[\n\\operatorname{det}\\left(\\begin{array}{ll}\n\\vec{v}_{1} \\cdot \\vec{v}_{1} & \\vec{v}_{1} \\cdot \\vec{v}_{2} \\\\\n\\vec{v}_{2} \\cdot \\vec{v}_{1} & \\vec{v}_{2} \\cdot \\vec{v}_{2}\n\\end{array}\\right)>0\n\\]\n(by the two-dimensional version of (3)) so the first two equations in (4) have a unique rational solution for \\( x, y \\), say \\( x=X / D, y=Y / D \\), where \\( X, Y \\), and \\( D \\) are integers. We may assume \\( \\operatorname{gcd}(X, Y, D)=1 \\). Then multiplying (4) through by \\( D \\) we have\n\\[\n\\begin{aligned}\nD & \\equiv 2 X+Y \\quad(\\bmod 8) \\\\\nD & \\equiv X+2 Y \\quad(\\bmod 8) \\\\\n2 D & \\equiv X+Y \\quad(\\bmod 8) .\n\\end{aligned}\n\\]\n\nAdding the first two congruences and subtracting the third gives \\( 2 X+2 Y \\equiv 0(\\bmod 8) \\), so, by the third congruence, \\( D \\) is even. But then the first two congruences force \\( X \\) and \\( Y \\) to be even, giving a contradiction.\n\nSolution 5. If \\( P_{0}, \\ldots, P_{n} \\) are the vertices of a simplex in \\( \\mathbb{R}^{n} \\), and \\( d_{i j}=\\left(P_{i} P_{j}\\right)^{2} \\), then its volume \\( V \\) satisfies\n\\[\nV^{2}=\\frac{(-1)^{n+1}}{2^{n}(n!)^{2}} \\operatorname{det}\\left(\\begin{array}{ccccc}\n0 & 1 & 1 & \\cdots & 1 \\\\\n1 & d_{00} & d_{01} & \\cdots & d_{0 n} \\\\\n1 & d_{10} & d_{11} & \\cdots & d_{1 n} \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n1 & d_{n 0} & d_{n 1} & \\cdots & d_{n n}\n\\end{array}\\right)\n\\]\n\nThis is proved on page 98 of [B1], where this determinant is called a Cayley-Menger determinant. In particular, the formula for \\( n=3 \\) (which goes back to Euler) implies that the edge lengths \\( a, b, c, x, y, z \\) (labelled as in Solution 2) of a degenerate tetrahedron satisfy\n\\[\n\\operatorname{det}\\left(\\begin{array}{ccccc}\n0 & 1 & 1 & 1 & 1 \\\\\n1 & 0 & z^{2} & y^{2} & a^{2} \\\\\n1 & z^{2} & 0 & x^{2} & b^{2} \\\\\n1 & y^{2} & x^{2} & 0 & c^{2} \\\\\n1 & a^{2} & b^{2} & c^{2} & 0\n\\end{array}\\right)=0 .\n\\]",
  "vars": [
    "D",
    "X",
    "Y",
    "d_ij",
    "i",
    "j",
    "r",
    "s",
    "t",
    "v_1",
    "v_2",
    "v_3",
    "v_i",
    "v_j",
    "x",
    "y",
    "z",
    "\\\\alpha",
    "\\\\beta"
  ],
  "params": [
    "A",
    "B",
    "C",
    "O",
    "P_0",
    "P_i",
    "P_n",
    "a",
    "b",
    "c",
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "D": "denomval",
        "X": "numervalx",
        "Y": "numervaly",
        "d_ij": "distij",
        "i": "indexi",
        "j": "indexj",
        "r": "cosfirst",
        "s": "cossecond",
        "t": "costhird",
        "v_1": "vectorone",
        "v_2": "vectortwo",
        "v_3": "vectorthree",
        "v_i": "vectori",
        "v_j": "vectorj",
        "x": "coordx",
        "y": "coordy",
        "z": "coordz",
        "\\alpha": "anglealpha",
        "\\beta": "anglebeta",
        "A": "pointa",
        "B": "pointb",
        "C": "pointc",
        "O": "pointo",
        "P_0": "pointzero",
        "P_i": "pointi",
        "P_n": "pointn",
        "a": "lengtha",
        "b": "lengthb",
        "c": "lengthc",
        "n": "dimcount"
      },
      "question": "Show there do not exist four points in the Euclidean plane such that the\npairwise distances between the points are all odd integers.",
      "solution": "Solution 1. For real numbers \\( coordx \\) and \\( coordy \\), and for a positive integer \\( dimcount \\), let \\\" \\( coordx \\equiv coordy \\) \\( (\\bmod dimcount) \\) \\\" mean that \\( coordx-coordy \\) is an integer divisible by \\( dimcount \\). Choose a coordinate system in which the four points are \\( (0,0),(lengtha, 0),(cosfirst, cossecond),(coordx, coordy) \\). Here \\( lengtha \\) is an odd integer, and we may assume \\( lengtha>0 \\). The square of an odd integer is congruent to 1 modulo 8, so if all the pairwise distances are odd integers, we have\n\\[\n\\begin{array}{l}\ncosfirst^{2}+cossecond^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(cosfirst-lengtha)^{2}+cossecond^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\ncoordx^{2}+coordy^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(coordx-lengtha)^{2}+coordy^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(coordx-cosfirst)^{2}+(coordy-cossecond)^{2} \\equiv 1 \\quad(\\bmod 8) .\n\\end{array}\n\\]\nSubtracting the first two yields \\( 2\\,lengtha\\,cosfirst \\equiv lengtha^{2}(\\bmod 8) \\). Thus \\( cosfirst \\) is a rational number whose denominator is a multiple of 2 and a divisor of \\( 2\\,lengtha \\). The same is true of \\( coordx \\). Therefore we can multiply all coordinates by the odd integer \\( lengtha \\), to reduce to the case where the denominators of \\( cosfirst \\) and \\( coordx \\) are both 2. Then the congruence \\( 2\\,lengtha\\,cosfirst \\equiv lengtha^{2}(\\bmod 8) \\) between integers implies \\( cosfirst \\equiv lengtha/2(\\bmod 4) \\). If \\( cosfirst=lengtha/2+4\\,lengthb \\), then\n\\[\ncosfirst^{2}=lengtha^{2}/4+4\\,lengtha\\,lengthb+16\\,lengthb^{2} \\equiv lengtha^{2}/4 \\quad(\\bmod 4),\n\\]\nso the first congruence implies\n\\[\ncossecond^{2} \\equiv 1-cosfirst^{2} \\equiv 1-lengtha^{2}/4 \\quad(\\bmod 4) .\n\\]\nSimilarly \\( coordx \\equiv lengtha/2(\\bmod 4) \\) and \\( coordy^{2} \\equiv 1-lengtha^{2}/4(\\bmod 4) \\). Also\n\\[\ncoordx-cosfirst \\equiv lengtha/2-lengtha/2 \\equiv 0 \\quad(\\bmod 4),\n\\]\nso \\( (coordx-cosfirst)^{2} \\in 16 \\mathbb{Z} \\), and the last of the five congruences yields\n\\[\n(coordy-cossecond)^{2} \\equiv 1-(coordx-cosfirst)^{2} \\equiv 1 \\quad(\\bmod 8).\n\\]\nWe will derive a contradiction from the congruences \\( cossecond^{2} \\equiv coordy^{2} \\equiv 1-lengtha^{2}/4(\\bmod 4) \\) and \\( (coordy-cossecond)^{2} \\equiv 1(\\bmod 8) \\) obtained above. First,\n\\[\n\\begin{aligned}\n(coordy+cossecond)^{2} & \\equiv 2\\,coordy^{2}+2\\,cossecond^{2}-(coordy-cossecond)^{2} \\\\\n& \\equiv 2\\left(1-lengtha^{2}/4\\right)+2\\left(1-lengtha^{2}/4\\right)-1 \\\\\n& \\equiv 3-lengtha^{2} \\\\\n& \\equiv 2 \\quad(\\bmod 4) .\n\\end{aligned}\n\\]\nMultiplying this integer congruence by \\( (coordy-cossecond)^{2} \\equiv 1(\\bmod 8) \\) yields \\( (coordy^{2}-cossecond^{2})^{2} \\equiv 2(\\bmod 4) \\). But \\( coordy^{2} \\) and \\( cossecond^{2} \\) are rational, so their difference is rational with square congruent to 2 modulo 4. This is impossible.\n\nSolution 2. Lemma. If \\( cosfirst=\\cos anglealpha,\\;cossecond=\\cos anglebeta \\), and \\( costhird=\\cos(anglealpha+anglebeta) \\), then \\( 1-cosfirst^{2}-cossecond^{2}-costhird^{2}+2\\,cosfirst\\,cossecond\\,costhird=0 \\).\nProof.\n\\[\n\\begin{aligned}\n\\cos(anglealpha+anglebeta)& =\\cos anglealpha\\,\\cos anglebeta-\\sin anglealpha\\,\\sin anglebeta \\\\\n(\\cos(anglealpha+anglebeta)-\\cos anglealpha\\,\\cos anglebeta)^{2} & =\\sin^{2} anglealpha\\,\\sin^{2} anglebeta \\\\\n(costhird-cosfirst\\,cossecond)^{2} & =(1-cosfirst^{2})(1-cossecond^{2}).\n\\end{aligned}\n\\]\nSuppose that \\( pointo, pointa, pointb, pointc \\) are four points in the plane such that \\( lengtha=pointo pointa,\\;lengthb=pointo pointb,\\;lengthc=pointo pointc,\\;coordx=pointb pointc,\\;coordy=pointc pointa,\\;coordz=pointa pointb \\) are all odd integers. Using the Law of Cosines, let\n\\[\n\\begin{array}{l}\ncosfirst=\\cos \\angle pointa pointo pointb=\\dfrac{lengtha^{2}+lengthb^{2}-coordz^{2}}{2\\,lengtha\\,lengthb} \\\\\ncossecond=\\cos \\angle pointb pointo pointc=\\dfrac{lengthb^{2}+lengthc^{2}-coordx^{2}}{2\\,lengthb\\,lengthc} \\\\\ncosthird=\\cos \\angle pointa pointo pointc=\\dfrac{lengthc^{2}+lengtha^{2}-coordy^{2}}{2\\,lengthc\\,lengtha} .\n\\end{array}\n\\]\nBecause \\( \\angle pointa pointo pointb+\\angle pointb pointo pointc=\\angle pointa pointo pointc \\), the lemma gives \\( 1-cosfirst^{2}-cossecond^{2}-costhird^{2}+2\\,cosfirst\\,cossecond\\,costhird=0 \\). Substituting and multiplying by \\( 4\\,lengtha^{2}\\,lengthb^{2}\\,lengthc^{2} \\) yields\n\\[\n\\begin{aligned}\n4\\,lengtha^{2}\\,lengthb^{2}\\,lengthc^{2}&-lengthc^{2}(lengtha^{2}+lengthb^{2}-coordz^{2})^{2}-lengtha^{2}(lengthb^{2}+lengthc^{2}-coordx^{2})^{2}\\\\\n&-lengthb^{2}(lengthc^{2}+lengtha^{2}-coordy^{2})^{2}\n+(lengtha^{2}+lengthb^{2}-coordz^{2})(lengthb^{2}+lengthc^{2}-coordx^{2})(lengthc^{2}+lengtha^{2}-coordy^{2})=0.\n\\end{aligned}\n\\]\nSince the square of an odd integer is 1 modulo 4,\n\\[\n4-1-1-1+1 \\equiv 0 \\quad(\\bmod 4),\n\\]\na contradiction.\n\nSolution 3 (Manjul Bhargava). Suppose \\( \\vec{vectorone},\\vec{vectortwo},\\vec{vectorthree} \\) are vectors in 3-space. The volume \\( V \\) of the parallelepiped they span is \\( |\\vec{vectorone}\\cdot(\\vec{vectortwo}\\times\\vec{vectorthree})|=|\\det M| \\), where \\( M=(\\vec{vectorone}\\,\\vec{vectortwo}\\,\\vec{vectorthree}) \\). Since \\( \\det M=\\det M^{T} \\),\n\\[\nV^{2}=\\det\\!\n\\begin{pmatrix}\n\\vec{vectorone}\\!\\cdot\\!\\vec{vectorone}&\\vec{vectorone}\\!\\cdot\\!\\vectortwo&\\vec{vectorone}\\!\\cdot\\!\\vectorthree\\\\\n\\vectortwo\\!\\cdot\\!\\vectorone&\\vectortwo\\!\\cdot\\!\\vectortwo&\\vectortwo\\!\\cdot\\!\\vectorthree\\\\\n\\vectorthree\\!\\cdot\\!\\vectorone&\\vectorthree\\!\\cdot\\!\\vectortwo&\\vectorthree\\!\\cdot\\!\\vectorthree\n\\end{pmatrix}.\n\\]\nThe volume of the tetrahedron they span is \\( V/6 \\). If its edges are \\( lengtha=|\\vec{vectorone}|,\\;lengthb=|\\vec{vectortwo}|,\\;lengthc=|\\vec{vectorthree}|,\\;coordx=|\\vec{vectortwo}-\\vec{vectorthree}|,\\;coordy=|\\vec{vectorthree}-\\vec{vectorone}|,\\;coordz=|\\vec{vectorone}-\\vec{vectortwo}| \\), then the Law of Cosines gives\n\\[\n2\\,\\vec{vectori}\\cdot\\vec{vectorj}=|\\vec{vectori}|^{2}+|\\vec{vectorj}|^{2}-|\\vec{vectori}-\\vec{vectorj}|^{2},\n\\]\nso\n\\[\n8V^{2}=\\det\\!\n\\begin{pmatrix}\n2\\,lengtha^{2}&lengtha^{2}+lengthb^{2}-coordz^{2}&lengtha^{2}+lengthc^{2}-coordy^{2}\\\\\nlengtha^{2}+lengthb^{2}-coordz^{2}&2\\,lengthb^{2}&lengthb^{2}+lengthc^{2}-coordx^{2}\\\\\nlengtha^{2}+lengthc^{2}-coordy^{2}&lengthb^{2}+lengthc^{2}-coordx^{2}&2\\,lengthc^{2}\n\\end{pmatrix}.\n\\]\nIf four points as in the problem existed, place one at the origin and let \\( \\vec{vectorone},\\vec{vectortwo},\\vec{vectorthree} \\) be the vectors to the others; they are coplanar, so \\( V=0 \\). Because squares of odd integers are 1 mod 8,\n\\[\n8V^{2}\\equiv \\det\\!\n\\begin{pmatrix}\n2&1&1\\\\1&2&1\\\\1&1&2\n\\end{pmatrix}\\equiv4\\pmod8,\n\\]\na contradiction.\n\nSolution 4. Assume again such 4 points exist and define \\( \\vec{vectorone},\\vec{vectortwo},\\vec{vectorthree} \\) as above. Then \\( \\vec{vectori}\\!\\cdot\\!\\vec{vectori}\\equiv1(\\bmod8) \\) and, for \\( indexi\\neq indexj \\), \\( 2\\,\\vec{vectori}\\!\\cdot\\!\\vec{vectorj}\\equiv1(\\bmod8) \\).\nNo three points are collinear, so \\( \\vec{vectorthree}=coordx\\,\\vec{vectorone}+coordy\\,\\vec{vectortwo} \\) for scalars \\( coordx,coordy \\). Hence\n\\[\n\\begin{array}{l}\n2\\,\\vec{vectorone}\\!\\cdot\\!\\vectorthree=2\\,coordx\\,\\vec{vectorone}\\!\\cdot\\!\\vectorone+2\\,coordy\\,\\vec{vectorone}\\!\\cdot\\!\\vectortwo\\\\\n2\\,\\vectortwo\\!\\cdot\\!\\vectorthree=2\\,coordx\\,\\vectortwo\\!\\cdot\\!\\vectorone+2\\,coordy\\,\\vectortwo\\!\\cdot\\!\\vectortwo\\\\\n2\\,\\vectorthree\\!\\cdot\\!\\vectorthree=2\\,coordx\\,\\vectorthree\\!\\cdot\\!\\vectorone+2\\,coordy\\,\\vectorthree\\!\\cdot\\!\\vectortwo.\n\\end{array}\n\\]\nSince \\( \\vec{vectorone} \\) is not a scalar multiple of \\( \\vec{vectortwo} \\),\n\\[\n\\det\\!\n\\begin{pmatrix}\n\\vec{vectorone}\\!\\cdot\\!\\vectorone&\\vec{vectorone}\\!\\cdot\\!\\vectortwo\\\\\n\\vectortwo\\!\\cdot\\!\\vectorone&\\vectortwo\\!\\cdot\\!\\vectortwo\n\\end{pmatrix}>0,\n\\]\nso the first two equations have a unique rational solution \\( coordx=numervalx/denomval,\\;coordy=numervaly/denomval \\) with \\( \\gcd(numervalx,numervaly,denomval)=1 \\). Multiplying by \\( denomval \\) gives\n\\[\n\\begin{aligned}\ndenomval&\\equiv2\\,numervalx+numervaly\\pmod8,\\\\\ndenomval&\\equiv numervalx+2\\,numervaly\\pmod8,\\\\\n2\\,denomval&\\equiv numervalx+numervaly\\pmod8.\n\\end{aligned}\n\\]\nAdding the first two congruences and subtracting the third yields \\( 2\\,numervalx+2\\,numervaly\\equiv0(\\bmod8) \\), so \\( denomval \\) is even; then the first two congruences force \\( numervalx \\) and \\( numervaly \\) to be even, contradicting \\( \\gcd(numervalx,numervaly,denomval)=1 \\).\n\nSolution 5. If \\( pointzero,\\ldots,pointn \\) are the vertices of a simplex in \\( \\mathbb{R}^{dimcount} \\) and \\( distij=(pointi\\,pointj)^{2} \\), then\n\\[\nV^{2}=\\frac{(-1)^{dimcount+1}}{2^{dimcount}(dimcount !)^{2}}\\det\\!\n\\begin{pmatrix}\n0&1&1&\\cdots&1\\\\\n1&d_{00}&d_{01}&\\cdots&d_{0\\,dimcount}\\\\\n1&d_{10}&d_{11}&\\cdots&d_{1\\,dimcount}\\\\\n\\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\\n1&d_{\\,dimcount0}&d_{\\,dimcount1}&\\cdots&d_{\\,dimcount\\,dimcount}\n\\end{pmatrix}.\n\\]\n(See Cayley–Menger determinant.) For \\( dimcount=3 \\) this implies that the edge lengths \\( lengtha,lengthb,lengthc,coordx,coordy,coordz \\) of a degenerate tetrahedron satisfy\n\\[\n\\det\\!\n\\begin{pmatrix}\n0&1&1&1&1\\\\\n1&0&coordz^{2}&coordy^{2}&lengtha^{2}\\\\\n1&coordz^{2}&0&coordx^{2}&lengthb^{2}\\\\\n1&coordy^{2}&coordx^{2}&0&lengthc^{2}\\\\\n1&lengtha^{2}&lengthb^{2}&lengthc^{2}&0\n\\end{pmatrix}=0,\n\\]\nwhich is impossible when every entry involving a length is 1 modulo 8. Hence no four planar points can have all pairwise distances odd."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "sandcastle",
        "B": "meadowlark",
        "C": "candlestick",
        "O": "riverbank",
        "P_0": "thunderbolt",
        "P_i": "silverfish",
        "P_n": "dragonfly",
        "a": "watermelon",
        "b": "whistlebox",
        "c": "gumdropper",
        "n": "goldthread",
        "D": "marigold",
        "X": "lighthouse",
        "Y": "strawberry",
        "d_ij": "porcupine",
        "j": "blackboard",
        "r": "pendulum",
        "s": "chocolate",
        "t": "woodpecker",
        "v_1": "daffodil",
        "v_2": "horsewhip",
        "v_3": "arrowroot",
        "v_i": "raincloud",
        "v_j": "tablespoon",
        "x": "buttercup",
        "y": "windswept",
        "z": "afterglow",
        "\\alpha": "chrysanthemum",
        "\\beta": "hummingbird"
      },
      "question": "Show there do not exist four points in the Euclidean plane such that the\npairwise distances between the points are all odd integers.",
      "solution": "Solution 1. For real numbers \\( x \\) and \\( y \\), and for a positive integer \\( goldthread \\), let \" \\( x \\equiv y \\) \\( (\\bmod goldthread) \\) \" mean that \\( x-y \\) is an integer divisible by \\( goldthread \\). Choose a coordinate system in which the four points are \\( (0,0),(watermelon, 0),(pendulum, chocolate),(buttercup, windswept) \\). Here \\( watermelon \\) is an odd integer, and we may assume \\( watermelon>0 \\). The square of an odd integer is congruent to 1 modulo 8 , so if all the pairwise distances are odd integers, we have\n\\[\n\\begin{array}{l}\npendulum^{2}+chocolate^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(pendulum-watermelon)^{2}+chocolate^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\nbuttercup^{2}+windswept^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(buttercup-watermelon)^{2}+windswept^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(buttercup-pendulum)^{2}+(windswept-chocolate)^{2} \\equiv 1 \\quad(\\bmod 8) .\n\\end{array}\n\\]\n\nSubtracting the first two yields \\( 2 watermelon pendulum \\equiv watermelon^{2}(\\bmod 8) \\). Thus \\( pendulum \\) is a rational number whose denominator is a multiple of 2 and a divisor of \\( 2 watermelon \\). The same is true of \\( buttercup \\). Therefore we can multiply all coordinates by the odd integer \\( watermelon \\), to reduce to the case where the denominators of \\( pendulum \\) and \\( buttercup \\) are both 2 . Then the congruence \\( 2 watermelon pendulum \\equiv watermelon^{2}(\\bmod 8) \\) between integers implies \\( pendulum \\equiv watermelon / 2(\\bmod 4) \\). If \\( pendulum=watermelon / 2+4 whistlebox \\), then\n\\[\npendulum^{2}=watermelon^{2} / 4+4 watermelon whistlebox+16 whistlebox^{2} \\equiv watermelon^{2} / 4 \\quad(\\bmod 4),\n\\]\nso the first congruence implies\n\\[\nchocolate^{2} \\equiv 1-pendulum^{2} \\equiv 1-watermelon^{2} / 4 \\quad(\\bmod 4) .\n\\]\n\nSimilarly \\( buttercup \\equiv watermelon / 2(\\bmod 4) \\) and \\( windswept^{2} \\equiv 1-watermelon^{2} / 4(\\bmod 4) \\). Also\n\\[\nbuttercup-pendulum \\equiv watermelon / 2-watermelon / 2 \\equiv 0 \\quad(\\bmod 4),\n\\]\nso \\( (buttercup-pendulum)^{2} \\in 16 \\mathbb{Z} \\), and the last of the five congruences yields\n\\[\n(windswept-chocolate)^{2} \\equiv 1-(buttercup-pendulum)^{2} \\equiv 1 \\quad(\\bmod 8)\n\\]\n\nWe will derive a contradiction from the congruences \\( chocolate^{2} \\equiv windswept^{2} \\equiv 1-watermelon^{2} / 4(\\bmod 4) \\) and \\( (windswept-chocolate)^{2} \\equiv 1(\\bmod 8) \\) obtained above. First,\n\\[\n\\begin{aligned}\n(windswept+chocolate)^{2} & \\equiv 2 windswept^{2}+2 chocolate^{2}-(windswept-chocolate)^{2} \\\\\n& \\equiv 2\\left(1-watermelon^{2} / 4\\right)+2\\left(1-watermelon^{2} / 4\\right)-1 \\\\\n& \\equiv 3-watermelon^{2} \\\\\n& \\equiv 2 \\quad(\\bmod 4) .\n\\end{aligned}\n\\]\n\nMultiplying this integer congruence by \\( (windswept-chocolate)^{2} \\equiv 1(\\bmod 8) \\) yields \\( \\left(windswept^{2}-chocolate^{2}\\right)^{2} \\equiv 2 \\) \\( (\\bmod 4) \\). But \\( windswept^{2} \\) and \\( chocolate^{2} \\) are rational by the beginning of this paragraph, so \\( windswept^{2}-chocolate^{2} \\) is a rational number with square congruent to 2 modulo 4 . This is impossible.\n\nSolution 2.\nLemma. If \\( pendulum=\\cos chrysanthemum, chocolate=\\cos hummingbird \\), and \\( woodpecker=\\cos (chrysanthemum+hummingbird) \\), then \\( 1-pendulum^{2}-chocolate^{2}-woodpecker^{2}+2 pendulum chocolate woodpecker=0 \\). Proof. We have\n\\[\n\\begin{aligned}\n\\cos (chrysanthemum+hummingbird) & =\\cos chrysanthemum \\cos hummingbird-\\sin chrysanthemum \\sin hummingbird \\\\\n(\\cos (chrysanthemum+hummingbird)-\\cos chrysanthemum \\cos hummingbird)^{2} & =\\sin ^{2} chrysanthemum \\sin ^{2} hummingbird \\\\\n(woodpecker-pendulum chocolate)^{2} & =(1-pendulum^{2})(1-chocolate^{2})\n\\end{aligned}\n\\]\nfrom which the result follows.\nSuppose that \\( riverbank, sandcastle, meadowlark, candlestick \\) are four points in the plane such that \\( watermelon=riverbank sandcastle, whistlebox=riverbank meadowlark, gumdropper=riverbank candlestick, buttercup=meadowlark candlestick, windswept=candlestick sandcastle, afterglow=sandcastle meadowlark \\) are all odd integers. Using the Law of Cosines, let\n\\[\n\\begin{array}{l}\npendulum=\\cos \\angle sandcastle riverbank meadowlark=\\dfrac{watermelon^{2}+whistlebox^{2}-afterglow^{2}}{2 watermelon whistlebox} \\\\\nchocolate=\\cos \\angle meadowlark riverbank candlestick=\\dfrac{whistlebox^{2}+gumdropper^{2}-buttercup^{2}}{2 whistlebox gumdropper} \\\\\nwoodpecker=\\cos \\angle sandcastle riverbank candlestick=\\dfrac{gumdropper^{2}+watermelon^{2}-windswept^{2}}{2 gumdropper watermelon} .\n\\end{array}\n\\]\n\nBut \\( \\angle sandcastle riverbank meadowlark+\\angle meadowlark riverbank candlestick=\\angle sandcastle riverbank candlestick \\) as directed angles, so the lemma implies \\( 1-pendulum^{2}-chocolate^{2}- woodpecker^{2}+2 pendulum chocolate woodpecker=0 \\). Substituting the values of \\( pendulum, chocolate, woodpecker \\), and multiplying by \\( 4 watermelon^{2} whistlebox^{2} gumdropper^{2} \\) yields\n\\[\n\\begin{aligned}\n4 watermelon^{2} whistlebox^{2} gumdropper^{2}-gumdropper^{2}\\left(watermelon^{2}+whistlebox^{2}-afterglow^{2}\\right)^{2} & -watermelon^{2}\\left(whistlebox^{2}+gumdropper^{2}-buttercup^{2}\\right)^{2}-whistlebox^{2}\\left(gumdropper^{2}+watermelon^{2}-windswept^{2}\\right)^{2} \\\\\n& +\\left(watermelon^{2}+whistlebox^{2}-afterglow^{2}\\right)\\left(whistlebox^{2}+gumdropper^{2}-buttercup^{2}\\right)\\left(gumdropper^{2}+watermelon^{2}-windswept^{2}\\right)=0 .\n\\end{aligned}\n\\]\n\nThe square of an odd integer is 1 modulo 4 , so we obtain\n\\[\n4-1-1-1+1 \\equiv 0 \\quad(\\bmod 4),\n\\]\na contradiction.\nSolution 3 (Manjul Bhargava). Suppose \\( \\vec{daffodil}, \\vec{horsewhip}, \\vec{arrowroot} \\) are vectors in 3 -space. Then the volume \\( V \\) of the parallelepiped spanned by the vectors is given by \\( |\\vec{daffodil} \\cdot(\\vec{horsewhip} \\times \\vec{arrowroot})|= |\\operatorname{det} M| \\), where \\( M=(\\vec{daffodil} \\vec{horsewhip} \\vec{arrowroot}) \\) is the \\( 3 \\times 3 \\) matrix with entries given by the vectors \\( \\vec{daffodil}, \\vec{horsewhip}, \\vec{arrowroot} \\). Since \\( \\operatorname{det} M=\\operatorname{det} M^{T} \\),\n\\[\nV^{2}=M \\cdot M^{T}=\\operatorname{det}\\left(\\begin{array}{ccc}\n\\vec{daffodil} \\cdot \\vec{daffodil} & \\vec{daffodil} \\cdot \\vec{horsewhip} & \\vec{daffodil} \\cdot \\vec{arrowroot} \\\\\n\\vec{horsewhip} \\cdot \\vec{daffodil} & \\vec{horsewhip} \\cdot \\vec{horsewhip} & \\vec{horsewhip} \\cdot \\vec{arrowroot} \\\\\n\\vec{arrowroot} \\cdot \\vec{daffodil} & \\vec{arrowroot} \\cdot \\vec{horsewhip} & \\vec{arrowroot} \\cdot \\vec{arrowroot}\n\\end{array}\\right)\n\\]\n\nThe volume of the tetrahedron spanned by the vectors is \\( V / 6 \\). If the edges of the tetrahedron are \\( watermelon=|\\vec{daffodil}|, whistlebox=|\\vec{horsewhip}|, gumdropper=|\\vec{arrowroot}|, buttercup=|\\vec{horsewhip}-\\vec{arrowroot}|, windswept=|\\vec{arrowroot}-\\vec{daffodil}|, afterglow=|\\vec{daffodil}-\\vec{horsewhip}| \\), then by the Law of Cosines, written as\n\\[\n\\begin{array}{c}\n2 \\vec{v}_{i} \\cdot \\vec{v}_{j}=|\\vec{v}_{i}|^{2}+|\\vec{v}_{j}|^{2}-|\\vec{v}_{i}-\\vec{v}_{j}|^{2}, \\\\\n8 V^{2}=\\operatorname{det}\\left(\\begin{array}{ccc}\n2 watermelon^{2} & watermelon^{2}+whistlebox^{2}-afterglow^{2} & watermelon^{2}+gumdropper^{2}-windswept^{2} \\\\\nwatermelon^{2}+whistlebox^{2}-afterglow^{2} & 2 whistlebox^{2} & whistlebox^{2}+gumdropper^{2}-buttercup^{2} \\\\\nwatermelon^{2}+gumdropper^{2}-windswept^{2} & whistlebox^{2}+gumdropper^{2}-buttercup^{2} & 2 gumdropper^{2}\n\\end{array}\\right) .\n\\end{array}\n\\]\n\nSuppose now that there were 4 points as described in the problem. Locate one point at the origin, and let \\( \\vec{daffodil}, \\vec{horsewhip}, \\vec{arrowroot} \\) be the vectors from the origin to the other three; they are coplanar, so \\( V=0 \\). Since squares of odd integers are congruent to 1 modulo 8 ,\n\\[\n8 V^{2} \\equiv \\operatorname{det}\\left(\\begin{array}{lll}\n2 & 1 & 1 \\\\\n1 & 2 & 1 \\\\\n1 & 1 & 2\n\\end{array}\\right) \\equiv 4 \\quad(\\bmod 8)\n\\]\nwhich gives a contradiction.\nSolution 4. Suppose as before there were 4 such points; define \\( \\vec{daffodil}, \\vec{horsewhip}, \\vec{arrowroot} \\) as in the previous solution. Then \\( \\vec{v}_{i} \\cdot \\vec{v}_{i} \\equiv 1(\\bmod 8) \\), and from \\( (2), 2 \\vec{v}_{i} \\cdot \\vec{v}_{j} \\equiv 1(\\bmod 8) \\) for \\( i \\neq j \\) as well.\nNo three of the points can be collinear. Hence \\( \\vec{arrowroot}=buttercup \\vec{daffodil}+windswept \\vec{horsewhip} \\) for some scalars \\( buttercup \\) and \\( windswept \\). Then\n\\[\n\\begin{array}{l}\n2 \\vec{daffodil} \\cdot \\vec{arrowroot}=2 buttercup \\vec{daffodil} \\cdot \\vec{daffodil}+2 windswept \\vec{daffodil} \\cdot \\vec{horsewhip} \\\\\n2 \\vec{horsewhip} \\cdot \\vec{arrowroot}=2 buttercup \\vec{horsewhip} \\cdot \\vec{daffodil}+2 windswept \\vec{horsewhip} \\cdot \\vec{horsewhip} \\\\\n2 \\vec{arrowroot} \\cdot \\vec{arrowroot}=2 buttercup \\vec{arrowroot} \\cdot \\vec{daffodil}+2 windswept \\vec{arrowroot} \\cdot \\vec{horsewhip} .\n\\end{array}\n\\]\n\nSince \\( \\vec{daffodil} \\) is not a scalar multiple of \\( \\vec{horsewhip} \\),\n\\[\n\\operatorname{det}\\left(\\begin{array}{ll}\n\\vec{daffodil} \\cdot \\vec{daffodil} & \\vec{daffodil} \\cdot \\vec{horsewhip} \\\\\n\\vec{horsewhip} \\cdot \\vec{daffodil} & \\vec{horsewhip} \\cdot \\vec{horsewhip}\n\\end{array}\\right)>0\n\\]\n(by the two-dimensional version of (3)) so the first two equations in (4) have a unique rational solution for \\( buttercup, windswept \\), say \\( buttercup=lighthouse / marigold, windswept=strawberry / marigold \\), where \\( lighthouse, strawberry, marigold \\) are integers. We may assume \\( \\operatorname{gcd}(lighthouse, strawberry, marigold)=1 \\). Then multiplying (4) through by \\( marigold \\) we have\n\\[\n\\begin{aligned}\nmarigold & \\equiv 2 lighthouse+strawberry \\quad(\\bmod 8) \\\\\nmarigold & \\equiv lighthouse+2 strawberry \\quad(\\bmod 8) \\\\\n2 marigold & \\equiv lighthouse+strawberry \\quad(\\bmod 8) .\n\\end{aligned}\n\\]\n\nAdding the first two congruences and subtracting the third gives \\( 2 lighthouse+2 strawberry \\equiv 0(\\bmod 8) \\), so, by the third congruence, \\( marigold \\) is even. But then the first two congruences force \\( lighthouse \\) and \\( strawberry \\) to be even, giving a contradiction.\n\nSolution 5. If \\( thunderbolt, \\ldots, dragonfly \\) are the vertices of a simplex in \\( \\mathbb{R}^{goldthread} \\), and \\( porcupine=\\left(P_{i} P_{j}\\right)^{2} \\), then its volume \\( V \\) satisfies\n\\[\nV^{2}=\\frac{(-1)^{goldthread+1}}{2^{goldthread}(goldthread!)^{2}} \\operatorname{det}\\left(\\begin{array}{ccccc}\n0 & 1 & 1 & \\cdots & 1 \\\\\n1 & porcupine_{00} & porcupine_{01} & \\cdots & porcupine_{0 goldthread} \\\\\n1 & porcupine_{10} & porcupine_{11} & \\cdots & porcupine_{1 goldthread} \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n1 & porcupine_{goldthread 0} & porcupine_{goldthread 1} & \\cdots & porcupine_{goldthread goldthread}\n\\end{array}\\right)\n\\]\n\nThis is proved on page 98 of [B1], where this determinant is called a Cayley-Menger determinant. In particular, the formula for \\( goldthread=3 \\) (which goes back to Euler) implies that the edge lengths \\( watermelon, whistlebox, gumdropper, buttercup, windswept, afterglow \\) (labelled as in Solution 2) of a degenerate tetrahedron satisfy\n\\[\n\\operatorname{det}\\left(\\begin{array}{ccccc}\n0 & 1 & 1 & 1 & 1 \\\\\n1 & 0 & afterglow^{2} & windswept^{2} & watermelon^{2} \\\\\n1 & afterglow^{2} & 0 & buttercup^{2} & whistlebox^{2} \\\\\n1 & windswept^{2} & buttercup^{2} & 0 & gumdropper^{2} \\\\\n1 & watermelon^{2} & whistlebox^{2} & gumdropper^{2} & 0\n\\end{array}\\right)=0 .\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "D": "numeratorr",
        "X": "smallness",
        "Y": "lownesss",
        "d_ij": "closeness",
        "i": "wholeindex",
        "j": "entireindex",
        "r": "sinevalue",
        "s": "tangentval",
        "t": "versineval",
        "v_1": "staticone",
        "v_2": "statictwo",
        "v_3": "staticthree",
        "v_i": "staticindex",
        "v_j": "staticsub",
        "x": "stagnance",
        "y": "restfulness",
        "z": "immobility",
        "\\alpha": "omegaangle",
        "\\beta": "gammavalue",
        "A": "voidness",
        "B": "blankness",
        "C": "emptiness",
        "O": "infinitept",
        "P_0": "endpoint",
        "P_i": "endindex",
        "P_n": "endtotal",
        "a": "minuscule",
        "b": "tinyvalue",
        "c": "petitval",
        "n": "zerocount"
      },
      "question": "Show there do not exist four points in the Euclidean plane such that the\npairwise distances between the points are all odd integers.",
      "solution": "Solution 1. For real numbers \\( stagnance \\) and \\( restfulness \\), and for a positive integer \\( zerocount \\), let \" \\( stagnance \\equiv restfulness \\) \\((\\bmod zerocount)\\) \" mean that \\( stagnance-restfulness \\) is an integer divisible by \\( zerocount \\). Choose a coordinate system in which the four points are \\( (0,0),(minuscule, 0),(sinevalue, tangentval),(stagnance, restfulness) \\). Here \\( minuscule \\) is an odd integer, and we may assume \\( minuscule>0 \\). The square of an odd integer is congruent to 1 modulo 8 , so if all the pairwise distances are odd integers, we have\n\\[\n\\begin{array}{l}\nsinevalue^{2}+tangentval^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(sinevalue-minuscule)^{2}+tangentval^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\nstagnance^{2}+restfulness^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(stagnance-minuscule)^{2}+restfulness^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(stagnance-sinevalue)^{2}+(restfulness-tangentval)^{2} \\equiv 1 \\quad(\\bmod 8) .\n\\end{array}\n\\]\n\nSubtracting the first two yields \\( 2 minuscule sinevalue \\equiv minuscule^{2}(\\bmod 8) \\). Thus \\( sinevalue \\) is a rational number whose denominator is a multiple of 2 and a divisor of \\( 2 minuscule \\). The same is true of \\( stagnance \\). Therefore we can multiply all coordinates by the odd integer \\( minuscule \\), to reduce to the case where the denominators of \\( sinevalue \\) and \\( stagnance \\) are both 2 . Then the congruence \\( 2 minuscule sinevalue \\equiv minuscule^{2}(\\bmod 8) \\) between integers implies \\( sinevalue \\equiv minuscule / 2(\\bmod 4) \\). If \\( sinevalue=minuscule / 2+4 tinyvalue \\), then\n\\[\nsinevalue^{2}=minuscule^{2} / 4+4 minuscule tinyvalue+16 tinyvalue^{2} \\equiv minuscule^{2} / 4 \\quad(\\bmod 4),\n\\]\nso the first congruence implies\n\\[\ntangentval^{2} \\equiv 1-sinevalue^{2} \\equiv 1-minuscule^{2} / 4 \\quad(\\bmod 4) .\n\\]\n\nSimilarly \\( stagnance \\equiv minuscule / 2(\\bmod 4) \\) and \\( restfulness^{2} \\equiv 1-minuscule^{2} / 4(\\bmod 4) \\). Also\n\\[\nstagnance-sinevalue \\equiv minuscule / 2-minuscule / 2 \\equiv 0 \\quad(\\bmod 4),\n\\]\nso \\( (stagnance-sinevalue)^{2} \\in 16 \\mathbb{Z} \\), and the last of the five congruences yields\n\\[\n(restfulness-tangentval)^{2} \\equiv 1-(stagnance-sinevalue)^{2} \\equiv 1 \\quad(\\bmod 8)\n\\]\n\nWe will derive a contradiction from the congruences \\( tangentval^{2} \\equiv restfulness^{2} \\equiv 1-minuscule^{2} / 4(\\bmod 4) \\) and \\( (restfulness-tangentval)^{2} \\equiv 1(\\bmod 8) \\) obtained above. First,\n\\[\n\\begin{aligned}\n(restfulness+tangentval)^{2} & \\equiv 2 restfulness^{2}+2 tangentval^{2}-(restfulness-tangentval)^{2} \\\\\n& \\equiv 2\\left(1-minuscule^{2} / 4\\right)+2\\left(1-minuscule^{2} / 4\\right)-1 \\\\\n& \\equiv 3-minuscule^{2} \\\\\n& \\equiv 2 \\quad(\\bmod 4) .\n\\end{aligned}\n\\]\n\nMultiplying this integer congruence by \\( (restfulness-tangentval)^{2} \\equiv 1(\\bmod 8) \\) yields \\( \\left(restfulness^{2}-tangentval^{2}\\right)^{2} \\equiv 2 \\) \\( (\\bmod 4) \\). But \\( restfulness^{2} \\) and \\( tangentval^{2} \\) are rational by the beginning of this paragraph, so \\( restfulness^{2}-tangentval^{2} \\) is a rational number with square congruent to 2 modulo 4 . This is impossible.\n\nSolution 2.\nLemma. If \\( sinevalue=\\cos omegaangle, tangentval=\\cos gammavalue \\), and \\( versineval=\\cos (omegaangle+gammavalue) \\), then \\( 1-sinevalue^{2}-tangentval^{2}-versineval^{2}+2 sinevalue tangentval versineval=0 \\). Proof. We have\n\\[\n\\begin{aligned}\n\\cos (omegaangle+gammavalue) & =\\cos omegaangle \\cos gammavalue-\\sin omegaangle \\sin gammavalue \\\\\n(\\cos (omegaangle+gammavalue)-\\cos omegaangle \\cos gammavalue)^{2} & =\\sin ^{2} omegaangle \\sin ^{2} gammavalue \\\\\n(versineval-sinevalue tangentval)^{2} & =\\left(1-sinevalue^{2}\\right)\\left(1-tangentval^{2}\\right)\n\\end{aligned}\n\\]\nfrom which the result follows.\nSuppose that \\( infinitept, voidness, blankness, emptiness \\) are four points in the plane such that \\( minuscule=infinitept voidness, tinyvalue=infinitept blankness \\), \\( petitval=infinitept emptiness, stagnance=blankness emptiness, restfulness=emptiness voidness, immobility=voidness blankness \\) are all odd integers. Using the Law of Cosines, let\n\\[\n\\begin{array}{l}\nsinevalue=\\cos \\angle voidness infinitept blankness=\\frac{minuscule^{2}+tinyvalue^{2}-immobility^{2}}{2 minuscule tinyvalue} \\\\\ntangentval=\\cos \\angle blankness infinitept emptiness=\\frac{tinyvalue^{2}+petitval^{2}-stagnance^{2}}{2 tinyvalue petitval} \\\\\nversineval=\\cos \\angle voidness infinitept emptiness=\\frac{petitval^{2}+minuscule^{2}-restfulness^{2}}{2 petitval minuscule} .\n\\end{array}\n\\]\n\nBut \\( \\angle voidness infinitept blankness+\\angle blankness infinitept emptiness=\\angle voidness infinitept emptiness \\) as directed angles, so the lemma implies \\( 1-sinevalue^{2}-tangentval^{2}- versineval^{2}+2 sinevalue tangentval versineval=0 \\). Substituting the values of \\( sinevalue, tangentval, versineval \\), and multiplying by \\( 4 minuscule^{2} tinyvalue^{2} petitval^{2} \\) yields\n\\[\n\\begin{aligned}\n4 minuscule^{2} tinyvalue^{2} petitval^{2}-petitval^{2}\\left(minuscule^{2}+tinyvalue^{2}-immobility^{2}\\right)^{2} & -minuscule^{2}\\left(tinyvalue^{2}+petitval^{2}-stagnance^{2}\\right)^{2}-tinyvalue^{2}\\left(petitval^{2}+minuscule^{2}-restfulness^{2}\\right)^{2} \\\\\n& +\\left(minuscule^{2}+tinyvalue^{2}-immobility^{2}\\right)\\left(tinyvalue^{2}+petitval^{2}-stagnance^{2}\\right)\\left(petitval^{2}+minuscule^{2}-restfulness^{2}\\right)=0 .\n\\end{aligned}\n\\]\n\nThe square of an odd integer is 1 modulo 4 , so we obtain\n\\[\n4-1-1-1+1 \\equiv 0 \\quad(\\bmod 4),\n\\]\na contradiction.\n\nSolution 3 (Manjul Bhargava). Suppose \\( \\vec{staticone}, \\vec{statictwo}, \\vec{staticthree} \\) are vectors in 3 -space. Then the volume \\( V \\) of the parallelepiped spanned by the vectors is given by \\( \\left|\\vec{staticone} \\cdot\\left(\\vec{statictwo} \\times \\vec{staticthree}\\right)\\right|= \\) \\( |\\operatorname{det} M| \\), where \\( M=\\left(\\vec{staticone} \\vec{statictwo} \\vec{staticthree}\\right) \\) is the \\( 3 \\times 3 \\) matrix with entries given by the vectors \\( \\vec{staticone} \\), \\( \\vec{statictwo}, \\vec{staticthree} \\). Since \\( \\operatorname{det} M=\\operatorname{det} M^{T} \\),\n\\[\nV^{2}=M \\cdot M^{T}=\\operatorname{det}\\left(\\begin{array}{ccc}\n\\vec{staticone} \\cdot \\vec{staticone} & \\vec{staticone} \\cdot \\vec{statictwo} & \\vec{staticone} \\cdot \\vec{staticthree} \\\\\n\\vec{statictwo} \\cdot \\vec{staticone} & \\vec{statictwo} \\cdot \\vec{statictwo} & \\vec{statictwo} \\cdot \\vec{staticthree} \\\\\n\\vec{staticthree} \\cdot \\vec{staticone} & \\vec{staticthree} \\cdot \\vec{statictwo} & \\vec{staticthree} \\cdot \\vec{staticthree}\n\\end{array}\\right)\n\\]\n\nThe volume of the tetrahedron spanned by the vectors is \\( V / 6 \\). If the edges of the tetrahedron are \\( minuscule=\\left|\\vec{staticone}\\right|, tinyvalue=\\left|\\vec{statictwo}\\right|, petitval=\\left|\\vec{staticthree}\\right|, stagnance=\\left|\\vec{statictwo}-\\vec{staticthree}\\right|, restfulness=\\left|\\vec{staticthree}-\\vec{staticone}\\right|, immobility=\\left|\\vec{staticone}-\\vec{statictwo}\\right| \\), then by the Law of Cosines, written as\n\\[\n\\begin{array}{c}\n2 \\vec{staticone} \\cdot \\vec{statictwo}=\\left|\\vec{staticone}\\right|^{2}+\\left|\\vec{statictwo}\\right|^{2}-\\left|\\vec{staticone}-\\vec{statictwo}\\right|^{2}, \\\\\n8 V^{2}=\\operatorname{det}\\left(\\begin{array}{ccc}\n2 minuscule^{2} & minuscule^{2}+tinyvalue^{2}-immobility^{2} & minuscule^{2}+petitval^{2}-restfulness^{2} \\\\\nminuscule^{2}+tinyvalue^{2}-immobility^{2} & 2 tinyvalue^{2} & tinyvalue^{2}+petitval^{2}-stagnance^{2} \\\\\nminuscule^{2}+petitval^{2}-restfulness^{2} & tinyvalue^{2}+petitval^{2}-stagnance^{2} & 2 petitval^{2}\n\\end{array}\\right) .\n\\end{array}\n\\]\n\nSuppose now that there were 4 points as described in the problem. Locate one point at the origin, and let \\( \\vec{staticone}, \\vec{statictwo}, \\vec{staticthree} \\) be the vectors from the origin to the other three; they are coplanar, so \\( V=0 \\). Since squares of odd integers are congruent to 1 modulo 8 ,\n\\[\n8 V^{2} \\equiv \\operatorname{det}\\left(\\begin{array}{lll}\n2 & 1 & 1 \\\\\n1 & 2 & 1 \\\\\n1 & 1 & 2\n\\end{array}\\right) \\equiv 4 \\quad(\\bmod 8)\n\\]\nwhich gives a contradiction.\n\nSolution 4. Suppose as before there were 4 such points; define \\( \\vec{staticone}, \\vec{statictwo}, \\vec{staticthree} \\) as in the previous solution. Then \\( \\vec{staticone} \\cdot \\vec{staticone} \\equiv 1(\\bmod 8) \\), and from \\( (2), 2 \\vec{staticone} \\cdot \\vec{statictwo} \\equiv 1(\\bmod 8) \\) for \\( wholeindex \\neq entireindex \\) as well.\nNo three of the points can be collinear. Hence \\( \\vec{staticthree}=stagnance \\vec{staticone}+restfulness \\vec{statictwo} \\) for some scalars \\( stagnance \\) and \\( restfulness \\). Then\n\\[\n\\begin{array}{l}\n2 \\vec{staticone} \\cdot \\vec{staticthree}=2 stagnance \\vec{staticone} \\cdot \\vec{staticone}+2 restfulness \\vec{staticone} \\cdot \\vec{statictwo} \\\\\n2 \\vec{statictwo} \\cdot \\vec{staticthree}=2 stagnance \\vec{statictwo} \\cdot \\vec{staticone}+2 restfulness \\vec{statictwo} \\cdot \\vec{statictwo} \\\\\n2 \\vec{staticthree} \\cdot \\vec{staticthree}=2 stagnance \\vec{staticthree} \\cdot \\vec{staticone}+2 restfulness \\vec{staticthree} \\cdot \\vec{statictwo} .\n\\end{array}\n\\]\n\nSince \\( \\vec{staticone} \\) is not a scalar multiple of \\( \\vec{statictwo} \\),\n\\[\n\\operatorname{det}\\left(\\begin{array}{ll}\n\\vec{staticone} \\cdot \\vec{staticone} & \\vec{staticone} \\cdot \\vec{statictwo} \\\\\n\\vec{statictwo} \\cdot \\vec{staticone} & \\vec{statictwo} \\cdot \\vec{statictwo}\n\\end{array}\\right)>0\n\\]\n(by the two-dimensional version of (3)) so the first two equations in (4) have a unique rational solution for \\( stagnance, restfulness \\), say \\( stagnance=smallness / numeratorr, restfulness=lownesss / numeratorr \\), where \\( smallness, lownesss \\), and \\( numeratorr \\) are integers. We may assume \\( \\operatorname{gcd}(smallness, lownesss, numeratorr)=1 \\). Then multiplying (4) through by \\( numeratorr \\) we have\n\\[\n\\begin{aligned}\nnumeratorr & \\equiv 2 smallness+lownesss \\quad(\\bmod 8) \\\\\nnumeratorr & \\equiv smallness+2 lownesss \\quad(\\bmod 8) \\\\\n2 numeratorr & \\equiv smallness+lownesss \\quad(\\bmod 8) .\n\\end{aligned}\n\\]\n\nAdding the first two congruences and subtracting the third gives \\( 2 smallness+2 lownesss \\equiv 0(\\bmod 8) \\), so, by the third congruence, \\( numeratorr \\) is even. But then the first two congruences force \\( smallness \\) and \\( lownesss \\) to be even, giving a contradiction.\n\nSolution 5. If \\( endpoint, \\ldots, endtotal \\) are the vertices of a simplex in \\( \\mathbb{R}^{zerocount} \\), and \\( closeness_{wholeindex entireindex}=\\left(endpoint_{wholeindex} endpoint_{entireindex}\\right)^{2} \\), then its volume \\( V \\) satisfies\n\\[\nV^{2}=\\frac{(-1)^{zerocount+1}}{2^{zerocount}(zerocount!)^{2}} \\operatorname{det}\\left(\\begin{array}{ccccc}\n0 & 1 & 1 & \\cdots & 1 \\\\\n1 & closeness_{00} & closeness_{01} & \\cdots & closeness_{0\\zerocount} \\\\\n1 & closeness_{10} & closeness_{11} & \\cdots & closeness_{1\\zerocount} \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n1 & closeness_{\\zerocount 0} & closeness_{\\zerocount 1} & \\cdots & closeness_{\\zerocount \\zerocount}\n\\end{array}\\right)\n\\]\n\nThis is proved on page 98 of [B1], where this determinant is called a Cayley-Menger determinant. In particular, the formula for \\( zerocount=3 \\) (which goes back to Euler) implies that the edge lengths \\( minuscule, tinyvalue, petitval, stagnance, restfulness, immobility \\) (labelled as in Solution 2) of a degenerate tetrahedron satisfy\n\\[\n\\operatorname{det}\\left(\\begin{array}{ccccc}\n0 & 1 & 1 & 1 & 1 \\\\\n1 & 0 & immobility^{2} & restfulness^{2} & minuscule^{2} \\\\\n1 & immobility^{2} & 0 & stagnance^{2} & tinyvalue^{2} \\\\\n1 & restfulness^{2} & stagnance^{2} & 0 & petitval^{2} \\\\\n1 & minuscule^{2} & tinyvalue^{2} & petitval^{2} & 0\n\\end{array}\\right)=0 .\n"
    },
    "garbled_string": {
      "map": {
        "D": "qzxwvtnp",
        "X": "hjgrksla",
        "Y": "plqmskdt",
        "d_ij": "fnbhdjks",
        "i": "vcmtrplq",
        "j": "skdjbvma",
        "r": "bqplsdkt",
        "s": "fjslqmrt",
        "t": "gnskdjpl",
        "v_1": "htrbndsq",
        "v_2": "kqpldmsr",
        "v_3": "splndkhr",
        "v_i": "jdnwqpsl",
        "v_j": "rmqskdhf",
        "x": "ljtksown",
        "y": "pktmslgh",
        "z": "qmvlbnst",
        "\\\\alpha": "zxcmnbqp",
        "\\\\beta": "khgfdsal",
        "A": "xcvbzlaq",
        "B": "plokmnjh",
        "C": "awsedrft",
        "O": "qazwsxed",
        "P_0": "jklpoiuy",
        "P_i": "mnbvcxzq",
        "P_n": "rfvtgbyh",
        "a": "sdfghjkl",
        "b": "qwertyui",
        "c": "ghjklmnb",
        "n": "vbnmlkjh"
      },
      "question": "Show there do not exist four points in the Euclidean plane such that the pairwise distances between the points are all odd integers.",
      "solution": "Solution 1. For real numbers \\( ljtksown \\) and \\( pktmslgh \\), and for a positive integer \\( vbnmlkjh \\), let \" \\( ljtksown \\equiv pktmslgh \\) \\( (\\bmod vbnmlkjh) \\) \" mean that \\( ljtksown-pktmslgh \\) is an integer divisible by \\( vbnmlkjh \\). Choose a coordinate system in which the four points are \\( (0,0),(sdfghjkl, 0),(bqplsdkt, fjslqmrt),(ljtksown, pktmslgh) \\). Here \\( sdfghjkl \\) is an odd integer, and we may assume \\( sdfghjkl>0 \\). The square of an odd integer is congruent to 1 modulo 8, so if all the pairwise distances are odd integers, we have\n\\[\n\\begin{array}{l}\nbqplsdkt^{2}+fjslqmrt^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(bqplsdkt-sdfghjkl)^{2}+fjslqmrt^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\nljtksown^{2}+pktmslgh^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(ljtksown-sdfghjkl)^{2}+pktmslgh^{2} \\equiv 1 \\quad(\\bmod 8) \\\\\n(ljtksown-bqplsdkt)^{2}+(pktmslgh-fjslqmrt)^{2} \\equiv 1 \\quad(\\bmod 8) .\n\\end{array}\n\\]\nSubtracting the first two yields \\( 2 sdfghjkl bqplsdkt \\equiv sdfghjkl^{2}(\\bmod 8) \\). Thus \\( bqplsdkt \\) is a rational number whose denominator is a multiple of 2 and a divisor of \\( 2 sdfghjkl \\). The same is true of \\( ljtksown \\). Therefore we can multiply all coordinates by the odd integer \\( sdfghjkl \\), to reduce to the case where the denominators of \\( bqplsdkt \\) and \\( ljtksown \\) are both 2. Then the congruence \\( 2 sdfghjkl bqplsdkt \\equiv sdfghjkl^{2}(\\bmod 8) \\) between integers implies \\( bqplsdkt \\equiv sdfghjkl / 2(\\bmod 4) \\). If \\( bqplsdkt=sdfghjkl / 2+4 qwertyui \\), then\n\\[\nbqplsdkt^{2}=sdfghjkl^{2} / 4+4 sdfghjkl qwertyui+16 qwertyui^{2} \\equiv sdfghjkl^{2} / 4 \\quad(\\bmod 4),\n\\]\nso the first congruence implies\n\\[\nfjslqmrt^{2} \\equiv 1-bqplsdkt^{2} \\equiv 1-sdfghjkl^{2} / 4 \\quad(\\bmod 4) .\n\\]\nSimilarly \\( ljtksown \\equiv sdfghjkl / 2(\\bmod 4) \\) and \\( pktmslgh^{2} \\equiv 1-sdfghjkl^{2} / 4(\\bmod 4) \\). Also\n\\[\nljtksown-bqplsdkt \\equiv sdfghjkl / 2-sdfghjkl / 2 \\equiv 0 \\quad(\\bmod 4),\n\\]\nso \\( (ljtksown-bqplsdkt)^{2} \\in 16 \\mathbb{Z} \\), and the last of the five congruences yields\n\\[\n(pktmslgh-fjslqmrt)^{2} \\equiv 1-(ljtksown-bqplsdkt)^{2} \\equiv 1 \\quad(\\bmod 8)\n\\]\nWe will derive a contradiction from the congruences \\( fjslqmrt^{2} \\equiv pktmslgh^{2} \\equiv 1-sdfghjkl^{2} / 4(\\bmod 4) \\) and \\( (pktmslgh-fjslqmrt)^{2} \\equiv 1(\\bmod 8) \\) obtained above. First,\n\\[\n\\begin{aligned}\n(pktmslgh+fjslqmrt)^{2} & \\equiv 2 pktmslgh^{2}+2 fjslqmrt^{2}-(pktmslgh-fjslqmrt)^{2} \\\\\n& \\equiv 2\\left(1-sdfghjkl^{2} / 4\\right)+2\\left(1-sdfghjkl^{2} / 4\\right)-1 \\\\\n& \\equiv 3-sdfghjkl^{2} \\\\\n& \\equiv 2 \\quad(\\bmod 4) .\n\\end{aligned}\n\\]\nMultiplying this integer congruence by \\( (pktmslgh-fjslqmrt)^{2} \\equiv 1(\\bmod 8) \\) yields \\( \\left(pktmslgh^{2}-fjslqmrt^{2}\\right)^{2} \\equiv 2 \\) \\( (\\bmod 4) \\). But \\( pktmslgh^{2} \\) and \\( fjslqmrt^{2} \\) are rational by the beginning of this paragraph, so \\( pktmslgh^{2}-fjslqmrt^{2} \\) is a rational number with square congruent to 2 modulo 4. This is impossible.\n\nSolution 2.\nLemma. If \\( bqplsdkt=\\cos zxcmnbqp, fjslqmrt=\\cos khgfdsal \\), and \\( gnskdjpl=\\cos (zxcmnbqp+khgfdsal) \\), then \\( 1-bqplsdkt^{2}-fjslqmrt^{2}-gnskdjpl^{2}+2 bqplsdkt fjslqmrt gnskdjpl=0 \\).\nProof. We have\n\\[\n\\begin{aligned}\n\\cos (zxcmnbqp+khgfdsal) & =\\cos zxcmnbqp \\cos khgfdsal-\\sin zxcmnbqp \\sin khgfdsal \\\\\n(\\cos (zxcmnbqp+khgfdsal)-\\cos zxcmnbqp \\cos khgfdsal)^{2} & =\\sin ^{2} zxcmnbqp \\sin ^{2} khgfdsal \\\\\n(gnskdjpl-bqplsdkt fjslqmrt)^{2} & =\\left(1-bqplsdkt^{2}\\right)\\left(1-fjslqmrt^{2}\\right)\n\\end{aligned}\n\\]\nfrom which the result follows.\nSuppose that \\( qazwsxed, xcvbzlaq, plokmnjh, awsedrft \\) are four points in the plane such that \\( sdfghjkl=qazwsxed xcvbzlaq, qwertyui=qazwsxed plokmnjh, ghjklmnb=qazwsxed awsedrft, ljtksown=plokmnjh awsedrft, pktmslgh=awsedrft xcvbzlaq, qmvlbnst=xcvbzlaq plokmnjh \\) are all odd integers. Using the Law of Cosines, let\n\\[\n\\begin{array}{l}\nbqplsdkt=\\cos \\angle xcvbzlaq qazwsxed plokmnjh=\\frac{sdfghjkl^{2}+qwertyui^{2}-qmvlbnst^{2}}{2 sdfghjkl qwertyui} \\\\\nfjslqmrt=\\cos \\angle plokmnjh qazwsxed awsedrft=\\frac{qwertyui^{2}+ghjklmnb^{2}-ljtksown^{2}}{2 qwertyui ghjklmnb} \\\\\ngnskdjpl=\\cos \\angle xcvbzlaq qazwsxed awsedrft=\\frac{ghjklmnb^{2}+sdfghjkl^{2}-pktmslgh^{2}}{2 ghjklmnb sdfghjkl} .\n\\end{array}\n\\]\nBut \\( \\angle xcvbzlaq qazwsxed plokmnjh+\\angle plokmnjh qazwsxed awsedrft=\\angle xcvbzlaq qazwsxed awsedrft \\) as directed angles, so the lemma implies \\( 1-bqplsdkt^{2}-fjslqmrt^{2}-gnskdjpl^{2}+2 bqplsdkt fjslqmrt gnskdjpl=0 \\). Substituting the values of \\( bqplsdkt, fjslqmrt, gnskdjpl \\), and multiplying by \\( 4 sdfghjkl^{2} qwertyui^{2} ghjklmnb^{2} \\) yields\n\\[\n\\begin{aligned}\n4 sdfghjkl^{2} qwertyui^{2} ghjklmnb^{2}-ghjklmnb^{2}\\left(sdfghjkl^{2}+qwertyui^{2}-qmvlbnst^{2}\\right)^{2} & -sdfghjkl^{2}\\left(qwertyui^{2}+ghjklmnb^{2}-ljtksown^{2}\\right)^{2}-qwertyui^{2}\\left(ghjklmnb^{2}+sdfghjkl^{2}-pktmslgh^{2}\\right)^{2} \\\\\n& +\\left(sdfghjkl^{2}+qwertyui^{2}-qmvlbnst^{2}\\right)\\left(qwertyui^{2}+ghjklmnb^{2}-ljtksown^{2}\\right)\\left(ghjklmnb^{2}+sdfghjkl^{2}-pktmslgh^{2}\\right)=0 .\n\\end{aligned}\n\\]\nThe square of an odd integer is 1 modulo 4, so we obtain\n\\[\n4-1-1-1+1 \\equiv 0 \\quad(\\bmod 4),\n\\]\na contradiction.\n\nSolution 3 (Manjul Bhargava). Suppose \\( \\vec{htrbndsq}, \\vec{kqpldmsr}, \\vec{splndkhr} \\) are vectors in 3-space. Then the volume \\( V \\) of the parallelepiped spanned by the vectors is given by \\( |\\vec{htrbndsq} \\cdot(\\vec{kqpldmsr} \\times \\vec{splndkhr})|=|\\operatorname{det} M| \\), where \\( M=(\\vec{htrbndsq} \\; \\vec{kqpldmsr} \\; \\vec{splndkhr}) \\) is the \\( 3 \\times 3 \\) matrix with entries given by the vectors. Since \\( \\operatorname{det} M=\\operatorname{det} M^{T} \\),\n\\[\nV^{2}=M \\cdot M^{T}=\\operatorname{det}\\left(\\begin{array}{ccc}\n\\vec{htrbndsq} \\cdot \\vec{htrbndsq} & \\vec{htrbndsq} \\cdot \\vec{kqpldmsr} & \\vec{htrbndsq} \\cdot \\vec{splndkhr} \\\\\n\\vec{kqpldmsr} \\cdot \\vec{htrbndsq} & \\vec{kqpldmsr} \\cdot \\vec{kqpldmsr} & \\vec{kqpldmsr} \\cdot \\vec{splndkhr} \\\\\n\\vec{splndkhr} \\cdot \\vec{htrbndsq} & \\vec{splndkhr} \\cdot \\vec{kqpldmsr} & \\vec{splndkhr} \\cdot \\vec{splndkhr}\n\\end{array}\\right)\n\\]\nThe volume of the tetrahedron spanned by the vectors is \\( V / 6 \\). If the edges of the tetrahedron are \\( sdfghjkl=|\\vec{htrbndsq}|, qwertyui=|\\vec{kqpldmsr}|, ghjklmnb=|\\vec{splndkhr}|, ljtksown=|\\vec{kqpldmsr}-\\vec{splndkhr}|, pktmslgh=|\\vec{splndkhr}-\\vec{htrbndsq}|, qmvlbnst=|\\vec{htrbndsq}-\\vec{kqpldmsr}| \\), then by the Law of Cosines, written as\n\\[\n\\begin{array}{c}\n2 \\vec{jdnwqpsl} \\cdot \\vec{rmqskdhf}=|\\vec{jdnwqpsl}|^{2}+|\\vec{rmqskdhf}|^{2}-|\\vec{jdnwqpsl}-\\vec{rmqskdhf}|^{2}, \\\\\n8 V^{2}=\\operatorname{det}\\left(\\begin{array}{ccc}\n2 sdfghjkl^{2} & sdfghjkl^{2}+qwertyui^{2}-qmvlbnst^{2} & sdfghjkl^{2}+ghjklmnb^{2}-pktmslgh^{2} \\\\\nsdfghjkl^{2}+qwertyui^{2}-qmvlbnst^{2} & 2 qwertyui^{2} & qwertyui^{2}+ghjklmnb^{2}-ljtksown^{2} \\\\\nsdfghjkl^{2}+ghjklmnb^{2}-pktmslgh^{2} & qwertyui^{2}+ghjklmnb^{2}-ljtksown^{2} & 2 ghjklmnb^{2}\n\\end{array}\\right) .\n\\end{array}\n\\]\nSuppose now that there were 4 points as described in the problem. Locate one point at the origin, and let \\( \\vec{htrbndsq}, \\vec{kqpldmsr}, \\vec{splndkhr} \\) be the vectors from the origin to the other three; they are coplanar, so \\( V=0 \\). Since squares of odd integers are congruent to 1 modulo 8,\n\\[\n8 V^{2} \\equiv \\operatorname{det}\\left(\\begin{array}{lll}\n2 & 1 & 1 \\\\\n1 & 2 & 1 \\\\\n1 & 1 & 2\n\\end{array}\\right) \\equiv 4 \\quad(\\bmod 8)\n\\]\nwhich gives a contradiction.\n\nSolution 4. Suppose as before there were 4 such points; define \\( \\vec{htrbndsq}, \\vec{kqpldmsr}, \\vec{splndkhr} \\) as in the previous solution. Then \\( \\vec{htrbndsq} \\cdot \\vec{htrbndsq} \\equiv 1(\\bmod 8) \\), and from (2), \\( 2 \\vec{htrbndsq} \\cdot \\vec{kqpldmsr} \\equiv 1(\\bmod 8) \\) for \\( vcmtrplq \\neq skdjbvma \\) as well.\nNo three of the points can be collinear. Hence \\( \\vec{splndkhr}=ljtksown \\vec{htrbndsq}+pktmslgh \\vec{kqpldmsr} \\) for some scalars \\( ljtksown \\) and \\( pktmslgh \\). Then\n\\[\n\\begin{array}{l}\n2 \\vec{htrbndsq} \\cdot \\vec{splndkhr}=2 ljtksown \\vec{htrbndsq} \\cdot \\vec{htrbndsq}+2 pktmslgh \\vec{htrbndsq} \\cdot \\vec{kqpldmsr} \\\\\n2 \\vec{kqpldmsr} \\cdot \\vec{splndkhr}=2 ljtksown \\vec{kqpldmsr} \\cdot \\vec{htrbndsq}+2 pktmslgh \\vec{kqpldmsr} \\cdot \\vec{kqpldmsr} \\\\\n2 \\vec{splndkhr} \\cdot \\vec{splndkhr}=2 ljtksown \\vec{splndkhr} \\cdot \\vec{htrbndsq}+2 pktmslgh \\vec{splndkhr} \\cdot \\vec{kqpldmsr} .\n\\end{array}\n\\]\nSince \\( \\vec{htrbndsq} \\) is not a scalar multiple of \\( \\vec{kqpldmsr} \\),\n\\[\n\\operatorname{det}\\left(\\begin{array}{ll}\n\\vec{htrbndsq} \\cdot \\vec{htrbndsq} & \\vec{htrbndsq} \\cdot \\vec{kqpldmsr} \\\\\n\\vec{kqpldmsr} \\cdot \\vec{htrbndsq} & \\vec{kqpldmsr} \\cdot \\vec{kqpldmsr}\n\\end{array}\\right)>0\n\\]\n(by the two-dimensional version of (3)) so the first two equations in (4) have a unique rational solution for \\( ljtksown, pktmslgh \\), say \\( ljtksown=hjgrksla / qzxwvtnp, pktmslgh=plqmskdt / qzxwvtnp \\), where \\( hjgrksla, plqmskdt, qzxwvtnp \\) are integers. We may assume \\( \\operatorname{gcd}(hjgrksla, plqmskdt, qzxwvtnp)=1 \\). Then multiplying (4) through by \\( qzxwvtnp \\) we have\n\\[\n\\begin{aligned}\nqzxwvtnp & \\equiv 2 hjgrksla+plqmskdt \\quad(\\bmod 8) \\\\\nqzxwvtnp & \\equiv hjgrksla+2 plqmskdt \\quad(\\bmod 8) \\\\\n2 qzxwvtnp & \\equiv hjgrksla+plqmskdt \\quad(\\bmod 8) .\n\\end{aligned}\n\\]\nAdding the first two congruences and subtracting the third gives \\( 2 hjgrksla+2 plqmskdt \\equiv 0(\\bmod 8) \\), so, by the third congruence, \\( qzxwvtnp \\) is even. But then the first two congruences force \\( hjgrksla \\) and \\( plqmskdt \\) to be even, giving a contradiction.\n\nSolution 5. If \\( jklpoiuy, \\ldots, rfvtgbyh \\) are the vertices of a simplex in \\( \\mathbb{R}^{vbnmlkjh} \\), and \\( fnbhdjks=\\left(P_{vcmtrplq} P_{skdjbvma}\\right)^{2} \\), then its volume \\( V \\) satisfies\n\\[\nV^{2}=\\frac{(-1)^{vbnmlkjh+1}}{2^{vbnmlkjh}(vbnmlkjh!)^{2}} \\operatorname{det}\\left(\\begin{array}{ccccc}\n0 & 1 & 1 & \\cdots & 1 \\\\\n1 & fnbhdjks_{00} & fnbhdjks_{01} & \\cdots & fnbhdjks_{0 vbnmlkjh} \\\\\n1 & fnbhdjks_{10} & fnbhdjks_{11} & \\cdots & fnbhdjks_{1 vbnmlkjh} \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n1 & fnbhdjks_{vbnmlkjh 0} & fnbhdjks_{vbnmlkjh 1} & \\cdots & fnbhdjks_{vbnmlkjh vbnmlkjh}\n\\end{array}\\right)\n\\]\nThis is proved on page 98 of [B1], where this determinant is called a Cayley-Menger determinant. In particular, the formula for \\( vbnmlkjh=3 \\) (which goes back to Euler) implies that the edge lengths \\( sdfghjkl, qwertyui, ghjklmnb, ljtksown, pktmslgh, qmvlbnst \\) (labelled as in Solution 2) of a degenerate tetrahedron satisfy\n\\[\n\\operatorname{det}\\left(\\begin{array}{ccccc}\n0 & 1 & 1 & 1 & 1 \\\\\n1 & 0 & qmvlbnst^{2} & pktmslgh^{2} & sdfghjkl^{2} \\\\\n1 & qmvlbnst^{2} & 0 & ljtksown^{2} & qwertyui^{2} \\\\\n1 & pktmslgh^{2} & ljtksown^{2} & 0 & ghjklmnb^{2} \\\\\n1 & sdfghjkl^{2} & qwertyui^{2} & ghjklmnb^{2} & 0\n\\end{array}\\right)=0 .\n"
    },
    "kernel_variant": {
      "question": "Show that there are no four points in the Euclidean plane whose six pair-wise distances are positive integers, each of which is congruent to 1 modulo 4.",
      "solution": "Assume, for the sake of contradiction, that such points A₁ ,A₂ ,A₃ ,A₄ do exist.\n\nStep 1 – Choice of coordinates\nBy a rigid motion place A₁=(0,1) and A₂=(b,1) with b>0 an odd integer satisfying b≡1 (mod 4).  Write A₃=(r,s) and A₄=(x,y).\n\nStep 2 – Working modulo 8\nBecause the square of every odd integer is 1 modulo 8, the six distance conditions give\n(i)  r²+(s−1)² ≡1 (mod 8),\n(ii) (r−b)²+(s−1)² ≡1 (mod 8),\n(iii) x²+(y−1)² ≡1 (mod 8),\n(iv) (x−b)²+(y−1)² ≡1 (mod 8),\n(v)  (x−r)²+(y−s)² ≡1 (mod 8).\n\nStep 3 – r and x are half–integers and satisfy 2r ≡ 2x ≡ b (mod 8)\nSubtract (i) from (ii):\n          (r−b)²−r² ≡0 (mod 8)\n⇔ −2br+b² ≡0 (mod 8)\n⇔ 2br ≡ b² (mod 8). (1)\nThe odd integer b is invertible modulo 8 (its inverse is b itself), so we may multiply (1) by b⁻¹ ≡ b (mod 8):\n          2r ≡ b (mod 8). (2)\nThus 2r is an integer, i.e. r∈½ℤ.  Repeating the same argument with x in place of r gives\n          2x ≡ b (mod 8). (3)\nHence x also lies in ½ℤ.\n\nMultiplying all coordinates by the odd integer b clears any odd denominators, so from now on we may treat r and x as half-integers while s and y are arbitrary reals.\n\nA refinement modulo 16.  Double congruence (1):\n          4br ≡ 2b² (mod 16).\nCancelling the odd factor b modulo 16 gives\n          4r ≡ 2b (mod 16) ⇒ 2r ≡ b (mod 8) (nothing new) ⇒ r = b/2 + 4k, k∈ℤ,\nso\n          r ≡ b/2 (mod 4). (4)\nExactly the same reasoning yields\n          x ≡ b/2 (mod 4). (5)\n\nStep 4 – Information about s and y\nInsert (4) into (i):\n          (s−1)² ≡ 1 − b²/4 (mod 4). (6)\nLikewise, using (iii) together with (5),\n          (y−1)² ≡ 1 − b²/4 (mod 4). (7)\nThus S²:=(s−1)² and Y²:=(y−1)² are rational numbers whose denominators divide 4.\n\nStep 5 – The long diagonal A₃A₄\nBecause of (4) and (5) we have x−r ≡0 (mod 4); hence (x−r)² is a multiple of 16.  Putting this into (v) and reducing modulo 8 gives\n          (y−s)² ≡1 (mod 8). (8)\n\nStep 6 – A contradiction\nPut S=s−1 and Y=y−1.  Congruences (6)–(8) read\n          S² ≡ Y² ≡ 1−b²/4 (mod 4), (9)\n          (Y−S)² ≡1 (mod 8). (10)\nCompute (Y+S)²:\n          (Y+S)² = 2Y²+2S²−(Y−S)²\n                   ≡ 2(1−b²/4)+2(1−b²/4)−1\n                   ≡ 3−b² ≡2 (mod 4) (b≡1 mod 4). (11)\nMultiplying (10) and (11):\n          (Y−S)²·(Y+S)² = (Y²−S²)² ≡1·2 ≡2 (mod 4). (12)\nWrite D = Y²−S².  From (9) we know S² and Y² have denominator dividing 4, so D = k/4 for some integer k.  Substitute into (12):\n          (k/4)² ≡2 (mod 4)\n⇔ k²/16 ≡2 (mod 4)\n⇔ k² ≡32 (mod 64). (13)\nBut the quadratic residues modulo 64 are\n0,1,4,9,16,17,25,33,36,41,49,57 – 32 is not among them.  Thus (13) is impossible.\n\nThis contradiction shows that our initial assumption was false: no four points in the Euclidean plane can have all six pairwise distances positive integers that are ≡1 (mod 4).",
      "_meta": {
        "core_steps": [
          "Place one point at (0,0) and another at (a,0) with a odd; call the other two (r,s) and (x,y).",
          "Because an odd square ≡1 (mod 8), write the five distance‐squared conditions as congruences ≡1 (mod 8).",
          "Subtract to get 2ar≡a² (mod 8); clear denominators so r,x have denominator 2, giving r,x≡a/2 (mod 4) and hence s²,y²≡1−a²/4 (mod 4).",
          "Note (x−r)²∈16ℤ, so (y−s)²≡1 (mod 8).",
          "Combine (y+s)² and (y−s)² to deduce (y²−s²)²≡2 (mod 4), impossible—contradiction."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Choice of the two points forced onto the x-axis (one at the origin, the other a units away). Any pair can be used.",
            "original": "(0,0) and (a,0)"
          },
          "slot2": {
            "description": "Numerical value of the baseline odd distance a; only oddness is needed, not its specific size.",
            "original": "a (unspecified positive odd integer)"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}