{
  "index": "1987-B-3",
  "type": "ALG",
  "tag": [
    "ALG",
    "GEO",
    "NT"
  ],
  "difficulty": "",
  "question": "Let $F$ be a field in which $1+1 \\neq 0$. Show that the set of\nsolutions to the equation $x^2+y^2=1$ with $x$ and $y$ in $F$ is given\nby $(x,y)=(1,0)$ and\n\\[\n(x,y) = \\left( \\frac{r^2-1}{r^2+1}, \\frac{2r}{r^2+1} \\right)\n\\]\nwhere $r$ runs through the elements of $F$ such that $r^2\\neq -1$.",
  "solution": "Solution 1. For \\( r^{2} \\neq-1 \\), let \\( \\left(x_{r}, y_{r}\\right)=\\left(\\frac{r^{2}-1}{r^{2}+1}, \\frac{2 r}{r^{2}+1}\\right) \\). Clearly \\( (1,0) \\) and \\( \\left(x_{r}, y_{r}\\right) \\) for \\( r^{2} \\neq-1 \\) are solutions.\n\nConversely suppose \\( x^{2}+y^{2}=1 \\). If \\( x=1 \\), then \\( y=0 \\) and we have \\( (1,0) \\). Otherwise define \\( r=y /(1-x) \\). (The reason for this is that \\( y_{r} /\\left(1-x_{r}\\right)=r \\).) Then \\( 1-x^{2}=y^{2}=r^{2}(1-x)^{2} \\), but \\( x \\neq 1 \\), so we may divide by \\( 1-x \\) to obtain \\( 1+x=r^{2}(1-x) \\), and \\( \\left(r^{2}+1\\right) x=\\left(r^{2}-1\\right) \\). If \\( r^{2}=-1 \\), then this says \\( 0=-2 \\), contradicting \\( 1+1 \\neq 0 \\). Thus \\( r^{2} \\neq-1, x=\\frac{r^{2}-1}{r^{2}+1}=x_{r} \\), and \\( y=r(1-x)=\\frac{2 r}{r^{2}+1}=y_{r} \\). Hence every solution to \\( x^{2}+y^{2}=1 \\) not equal to \\( (1,0) \\) is of the form \\( \\left(x_{r}, y_{r}\\right) \\) for some \\( r \\in F \\) with \\( r^{2} \\neq-1 \\).\n\nRemark. If instead \\( F \\) is a field in which \\( 1+1=0 \\) (i.e., the characteristic of \\( F \\) is 2), then \\( x^{2}+y^{2}=1 \\) is equivalent to \\( (x+y+1)^{2}=0 \\), and the set of solutions is \\( \\{(t, t+1): t \\in F\\} \\).\n\nSolution 2. Essentially the same solution can be motivated by geometry. The only solution to \\( x^{2}+y^{2}=1 \\) with \\( x=1 \\) is \\( (1,0) \\). For each \\( (x, y) \\in F^{2} \\) satisfying \\( x^{2}+y^{2}=1 \\) with \\( x \\neq 1 \\), the line through \\( (x, y) \\) and \\( (1,0) \\) has slope \\( y /(x-1) \\) in \\( F \\). Hence we can find all solutions to \\( x^{2}+y^{2}=1 \\) with \\( x \\neq 1 \\) by intersecting each nonvertical line through \\( (1,0) \\) with the circle. (See Figure 4.)\n\nIf \\( L_{s} \\) is the line through \\( (1,0) \\) with slope \\( s \\in F \\), its intersection with \\( x^{2}+y^{2}=1 \\) can be computed by substituting \\( y=s(x-1) \\) into \\( x^{2}+y^{2}=1 \\), and solving the resulting equation\n\\[\nx^{2}+s^{2}(x-1)^{2}=1\n\\]\n\nThis equation is guaranteed to have the solution \\( x=1 \\), because \\( (1,0) \\) is in the intersection. Therefore we obtain a factorization,\n\\[\n(x-1)\\left(\\left(s^{2}+1\\right) x+\\left(1-s^{2}\\right)\\right)=0\n\\]\nwhich yields the solutions \\( x=1 \\) and, if \\( s^{2} \\neq-1 \\), also \\( x=\\left(s^{2}-1\\right) /\\left(s^{2}+1\\right) \\). (If \\( s^{2}=-1 \\), then \\( 1-s^{2}=-2 \\neq 0 \\), and the second factor gives no solution.) Using \\( y=s(x-1) \\), we find that these give \\( (1,0) \\) and \\( \\left(\\frac{s^{2}-1}{s^{2}+1}, \\frac{-2 s}{s^{2}+1}\\right) \\), which, as we verify, do satisfy \\( x^{2}+y^{2}=1 \\) and \\( y=s(x-1) \\). We finish by substituting \\( s=-r \\).\n\nRemark. Let \\( C \\) be any nondegenerate conic over \\( F \\) with an \\( F \\)-rational point \\( P \\) : this means that \\( C \\) is given by a polynomial \\( f(x, y) \\in F[x, y] \\) of total degree 2 that does not factor into linear polynomials over any field extension of \\( F \\), and \\( P=(a, b) \\in F^{2} \\) is a point such that \\( f(a, b)=0 \\). The same method (of drawing all lines through \\( P \\) with slope in \\( F \\), and seeing where they intersect \\( f(x, y)=0 \\) other than at \\( P \\) ) lets one parameterize the set of \\( F \\)-rational points of \\( C \\) in terms of a single parameter \\( s \\). In the language of algebraic geometry, one says that any conic over \\( k \\) with a \\( k \\)-rational point is birationally equivalent to the line [Shaf, p. 11]. The same method works on certain equations of higher degree [NZM, Section 5.6].\n\nThe parameterization in the preceding paragraph is the reason why indefinite integrals of rational functions in \\( t \\) and \\( \\sqrt{p(t)} \\) for a single quadratic polynomial \\( p(t) \\) can be expressed in terms of elementary functions [Shaf, p. 7]. It also explains why \\( \\sin \\theta \\) and \\( \\cos \\theta \\) can be expressed as rational functions of a single function:\n\\[\n(\\cos \\theta, \\sin \\theta)=\\left(\\frac{1-t^{2}}{1+t^{2}}, \\frac{2 t}{1+t^{2}}\\right) \\quad \\text { where } t=\\tan (\\theta / 2)\n\\]\n\nThis, in turn, explains why rational functions in \\( \\sin \\theta \\) and \\( \\cos \\theta \\), such as\n\\[\n\\frac{\\sin ^{3} \\theta-7 \\sin \\theta \\cos \\theta}{1+\\cos ^{3} \\theta}\n\\]\nhave elementary antiderivatives.\nRemark. The parameterization of solutions to \\( x^{2}+y^{2}=1 \\) over \\( \\mathbb{Q} \\) is closely linked to the parameterization of primitive Pythagorean triples, i.e., positive integer solutions to \\( a^{2}+b^{2}=c^{2} \\) with \\( \\operatorname{gcd}(a, b, c)=1 \\). In any primitive Pythagorean triple, exactly one of \\( a \\) and \\( b \\) is even; the set of primitive Pythagorean triples \\( (a, b, c) \\) with \\( b \\) even equals the set of triples \\( \\left(m^{2}-n^{2}, 2 m n, m^{2}+n^{2}\\right) \\) with \\( m, n \\) ranging over positive integers of opposite parity satisfying \\( m>n \\) and \\( \\operatorname{gcd}(m, n)=1 \\). A more number-theoretic (less geometric) approach to this classification is given in [NZM, Section 5.3].",
  "vars": [
    "x",
    "x_r",
    "y",
    "y_r",
    "r",
    "s",
    "t",
    "a",
    "b",
    "c",
    "m",
    "n",
    "p",
    "\\\\theta",
    "L_s"
  ],
  "params": [
    "F",
    "k",
    "C",
    "P",
    "f"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "x_r": "paramxrow",
        "y": "variabley",
        "y_r": "paramyrow",
        "r": "paramrad",
        "s": "slopevar",
        "t": "paramtan",
        "a": "coeffalfa",
        "b": "coeffbeta",
        "c": "coeffchar",
        "m": "indexmvar",
        "n": "indexnvar",
        "p": "polyparm",
        "\\theta": "angletheta",
        "L_s": "lineslope",
        "F": "fieldbase",
        "k": "fieldkapp",
        "C": "conicset",
        "P": "pointbase",
        "f": "polyfunc"
      },
      "question": "Let $fieldbase$ be a field in which $1+1 \\neq 0$. Show that the set of\nsolutions to the equation $variablex^{2}+variabley^{2}=1$ with $variablex$ and $variabley$ in $fieldbase$ is given\nby $(variablex,variabley)=(1,0)$ and\n\\[\n(variablex,variabley) = \\left( \\frac{paramrad^{2}-1}{paramrad^{2}+1}, \\frac{2paramrad}{paramrad^{2}+1} \\right)\n\\]\nwhere $paramrad$ runs through the elements of $fieldbase$ such that $paramrad^{2}\\neq -1$.",
      "solution": "Solution 1. For \\( paramrad^{2} \\neq -1 \\), let \\( \\left(paramxrow, paramyrow\\right)=\\left(\\frac{paramrad^{2}-1}{paramrad^{2}+1}, \\frac{2 paramrad}{paramrad^{2}+1}\\right) \\). Clearly \\( (1,0) \\) and \\( \\left(paramxrow, paramyrow\\right) \\) for \\( paramrad^{2} \\neq -1 \\) are solutions.\n\nConversely suppose \\( variablex^{2}+variabley^{2}=1 \\). If \\( variablex=1 \\), then \\( variabley=0 \\) and we have \\( (1,0) \\). Otherwise define \\( paramrad=variabley /(1-variablex) \\). (The reason for this is that \\( paramyrow /\\left(1-paramxrow\\right)=paramrad \\).) Then \\( 1-variablex^{2}=variabley^{2}=paramrad^{2}(1-variablex)^{2} \\), but \\( variablex \\neq 1 \\), so we may divide by \\( 1-variablex \\) to obtain \\( 1+variablex=paramrad^{2}(1-variablex) \\), and \\( \\left(paramrad^{2}+1\\right) variablex=\\left(paramrad^{2}-1\\right) \\). If \\( paramrad^{2}=-1 \\), then this says \\( 0=-2 \\), contradicting \\( 1+1 \\neq 0 \\). Thus \\( paramrad^{2} \\neq-1, variablex=\\frac{paramrad^{2}-1}{paramrad^{2}+1}=paramxrow \\), and \\( variabley=paramrad(1-variablex)=\\frac{2 paramrad}{paramrad^{2}+1}=paramyrow \\). Hence every solution to \\( variablex^{2}+variabley^{2}=1 \\) not equal to \\( (1,0) \\) is of the form \\( \\left(paramxrow, paramyrow\\right) \\) for some \\( paramrad \\in fieldbase \\) with \\( paramrad^{2} \\neq-1 \\).\n\nRemark. If instead \\( fieldbase \\) is a field in which \\( 1+1=0 \\) (i.e., the characteristic of \\( fieldbase \\) is 2), then \\( variablex^{2}+variabley^{2}=1 \\) is equivalent to \\( (variablex+variabley+1)^{2}=0 \\), and the set of solutions is \\( \\{(paramtan, paramtan+1): paramtan \\in fieldbase\\} \\).\n\nSolution 2. Essentially the same solution can be motivated by geometry. The only solution to \\( variablex^{2}+variabley^{2}=1 \\) with \\( variablex=1 \\) is \\( (1,0) \\). For each \\( (variablex, variabley) \\in fieldbase^{2} \\) satisfying \\( variablex^{2}+variabley^{2}=1 \\) with \\( variablex \\neq 1 \\), the line through \\( (variablex, variabley) \\) and \\( (1,0) \\) has slope \\( variabley /(variablex-1) \\) in \\( fieldbase \\). Hence we can find all solutions to \\( variablex^{2}+variabley^{2}=1 \\) with \\( variablex \\neq 1 \\) by intersecting each nonvertical line through \\( (1,0) \\) with the circle. (See Figure 4.)\n\nIf \\( lineslope \\) is the line through \\( (1,0) \\) with slope \\( slopevar \\in fieldbase \\), its intersection with \\( variablex^{2}+variabley^{2}=1 \\) can be computed by substituting \\( variabley=slopevar(variablex-1) \\) into \\( variablex^{2}+variabley^{2}=1 \\), and solving the resulting equation\n\\[\nvariablex^{2}+slopevar^{2}(variablex-1)^{2}=1\n\\]\nThis equation is guaranteed to have the solution \\( variablex=1 \\), because \\( (1,0) \\) is in the intersection. Therefore we obtain a factorization,\n\\[\n(variablex-1)\\left(\\left(slopevar^{2}+1\\right) variablex+\\left(1-slopevar^{2}\\right)\\right)=0\n\\]\nwhich yields the solutions \\( variablex=1 \\) and, if \\( slopevar^{2} \\neq-1 \\), also \\( variablex=\\left(slopevar^{2}-1\\right) /\\left(slopevar^{2}+1\\right) \\). (If \\( slopevar^{2}=-1 \\), then \\( 1-slopevar^{2}=-2 \\neq 0 \\), and the second factor gives no solution.) Using \\( variabley=slopevar(variablex-1) \\), we find that these give \\( (1,0) \\) and \\( \\left(\\frac{slopevar^{2}-1}{slopevar^{2}+1}, \\frac{-2 slopevar}{slopevar^{2}+1}\\right) \\), which, as we verify, do satisfy \\( variablex^{2}+variabley^{2}=1 \\) and \\( variabley=slopevar(variablex-1) \\). We finish by substituting \\( slopevar=-paramrad \\).\n\nRemark. Let \\( conicset \\) be any nondegenerate conic over \\( fieldbase \\) with an \\( fieldbase \\)-rational point \\( pointbase \\) : this means that \\( conicset \\) is given by a polynomial \\( polyfunc(variablex, variabley) \\in fieldbase[variablex, variabley] \\) of total degree 2 that does not factor into linear polynomials over any field extension of \\( fieldbase \\), and \\( pointbase=(coeffalfa, coeffbeta) \\in fieldbase^{2} \\) is a point such that \\( polyfunc(coeffalfa, coeffbeta)=0 \\). The same method (of drawing all lines through \\( pointbase \\) with slope in \\( fieldbase \\), and seeing where they intersect \\( polyfunc(variablex, variabley)=0 \\) other than at \\( pointbase \\) ) lets one parameterize the set of \\( fieldbase \\)-rational points of \\( conicset \\) in terms of a single parameter \\( slopevar \\). In the language of algebraic geometry, one says that any conic over \\( fieldkapp \\) with a \\( fieldkapp \\)-rational point is birationally equivalent to the line [Shaf, polyparm. 11]. The same method works on certain equations of higher degree [NZM, Section 5.6].\n\nThe parameterization in the preceding paragraph is the reason why indefinite integrals of rational functions in \\( paramtan \\) and \\( \\sqrt{polyparm(paramtan)} \\) for a single quadratic polynomial \\( polyparm(paramtan) \\) can be expressed in terms of elementary functions [Shaf, polyparm. 7]. It also explains why \\( \\sin angletheta \\) and \\( \\cos angletheta \\) can be expressed as rational functions of a single function:\n\\[\n(\\cos angletheta, \\sin angletheta)=\\left(\\frac{1-paramtan^{2}}{1+paramtan^{2}}, \\frac{2 paramtan}{1+paramtan^{2}}\\right) \\quad \\text { where } paramtan=\\tan (angletheta / 2)\n\\]\nThis, in turn, explains why rational functions in \\( \\sin angletheta \\) and \\( \\cos angletheta \\), such as\n\\[\n\\frac{\\sin ^{3} angletheta-7 \\sin angletheta \\cos angletheta}{1+\\cos ^{3} angletheta}\n\\]\nhave elementary antiderivatives.\nRemark. The parameterization of solutions to \\( variablex^{2}+variabley^{2}=1 \\) over \\( \\mathbb{Q} \\) is closely linked to the parameterization of primitive Pythagorean triples, i.e., positive integer solutions to \\( coeffalfa^{2}+coeffbeta^{2}=coeffchar^{2} \\) with \\( \\operatorname{gcd}(coeffalfa, coeffbeta, coeffchar)=1 \\). In any primitive Pythagorean triple, exactly one of \\( coeffalfa \\) and \\( coeffbeta \\) is even; the set of primitive Pythagorean triples \\( (coeffalfa, coeffbeta, coeffchar) \\) with \\( coeffbeta \\) even equals the set of triples \\( \\left(indexmvar^{2}-indexnvar^{2}, 2 indexmvar indexnvar, indexmvar^{2}+indexnvar^{2}\\right) \\) with \\( indexmvar, indexnvar \\) ranging over positive integers of opposite parity satisfying \\( indexmvar>indexnvar \\) and \\( \\operatorname{gcd}(indexmvar, indexnvar)=1 \\). A more number-theoretic (less geometric) approach to this classification is given in [NZM, Section 5.3]."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marzipans",
        "x_r": "driftwood",
        "y": "raincloud",
        "y_r": "lighthouse",
        "r": "quagmire",
        "s": "pinecones",
        "t": "drumstick",
        "a": "gravestone",
        "b": "honeycomb",
        "c": "peppercorn",
        "m": "shipwreck",
        "n": "snowflake",
        "p": "afterglow",
        "\\theta": "labyrinth",
        "L_s": "nightshade",
        "F": "sandstone",
        "k": "moonlight",
        "C": "sapphire",
        "P": "wildfire",
        "f": "compasses"
      },
      "question": "Let $sandstone$ be a field in which $1+1 \\neq 0$. Show that the set of\nsolutions to the equation $marzipans^2+raincloud^2=1$ with $marzipans$ and $raincloud$ in $sandstone$ is given\nby $(marzipans,raincloud)=(1,0)$ and\n\\[\n(marzipans,raincloud) = \\left( \\frac{quagmire^2-1}{quagmire^2+1}, \\frac{2quagmire}{quagmire^2+1} \\right)\n\\]\nwhere $quagmire$ runs through the elements of $sandstone$ such that $quagmire^2\\neq -1$.",
      "solution": "Solution 1. For \\( quagmire^{2} \\neq -1 \\), let \\( \\left(driftwood, lighthouse\\right)=\\left(\\frac{quagmire^{2}-1}{quagmire^{2}+1}, \\frac{2 quagmire}{quagmire^{2}+1}\\right) \\). Clearly \\( (1,0) \\) and \\( \\left(driftwood, lighthouse\\right) \\) for \\( quagmire^{2} \\neq -1 \\) are solutions.\n\nConversely suppose \\( marzipans^{2}+raincloud^{2}=1 \\). If \\( marzipans=1 \\), then \\( raincloud=0 \\) and we have \\( (1,0) \\). Otherwise define \\( quagmire=raincloud /(1-marzipans) \\). (The reason for this is that \\( lighthouse /\\left(1-driftwood\\right)=quagmire \\).) Then \\( 1-marzipans^{2}=raincloud^{2}=quagmire^{2}(1-marzipans)^{2} \\), but \\( marzipans \\neq 1 \\), so we may divide by \\( 1-marzipans \\) to obtain \\( 1+marzipans=quagmire^{2}(1-marzipans) \\), and \\( \\left(quagmire^{2}+1\\right) marzipans=\\left(quagmire^{2}-1\\right) \\). If \\( quagmire^{2}=-1 \\), then this says \\( 0=-2 \\), contradicting \\( 1+1 \\neq 0 \\). Thus \\( quagmire^{2} \\neq-1, marzipans=\\frac{quagmire^{2}-1}{quagmire^{2}+1}=driftwood \\), and \\( raincloud=quagmire(1-marzipans)=\\frac{2 quagmire}{quagmire^{2}+1}=lighthouse \\). Hence every solution to \\( marzipans^{2}+raincloud^{2}=1 \\) not equal to \\( (1,0) \\) is of the form \\( \\left(driftwood, lighthouse\\right) \\) for some \\( quagmire \\in sandstone \\) with \\( quagmire^{2} \\neq -1 \\).\n\nRemark. If instead \\( sandstone \\) is a field in which \\( 1+1=0 \\) (i.e., the characteristic of \\( sandstone \\) is 2), then \\( marzipans^{2}+raincloud^{2}=1 \\) is equivalent to \\( (marzipans+raincloud+1)^{2}=0 \\), and the set of solutions is \\( \\{(drumstick, drumstick+1): drumstick \\in sandstone\\} \\).\n\nSolution 2. Essentially the same solution can be motivated by geometry. The only solution to \\( marzipans^{2}+raincloud^{2}=1 \\) with \\( marzipans=1 \\) is \\( (1,0) \\). For each \\( (marzipans, raincloud) \\in sandstone^{2} \\) satisfying \\( marzipans^{2}+raincloud^{2}=1 \\) with \\( marzipans \\neq 1 \\), the line through \\( (marzipans, raincloud) \\) and \\( (1,0) \\) has slope \\( raincloud /(marzipans-1) \\) in \\( sandstone \\). Hence we can find all solutions to \\( marzipans^{2}+raincloud^{2}=1 \\) with \\( marzipans \\neq 1 \\) by intersecting each nonvertical line through \\( (1,0) \\) with the circle. (See Figure 4.)\n\nIf \\( nightshade \\) is the line through \\( (1,0) \\) with slope \\( pinecones \\in sandstone \\), its intersection with \\( marzipans^{2}+raincloud^{2}=1 \\) can be computed by substituting \\( raincloud=pinecones(marzipans-1) \\) into \\( marzipans^{2}+raincloud^{2}=1 \\), and solving the resulting equation\n\\[\nmarzipans^{2}+pinecones^{2}(marzipans-1)^{2}=1\n\\]\nThis equation is guaranteed to have the solution \\( marzipans=1 \\), because \\( (1,0) \\) is in the intersection. Therefore we obtain a factorization,\n\\[\n(marzipans-1)\\left(\\left(pinecones^{2}+1\\right) marzipans+\\left(1-pinecones^{2}\\right)\\right)=0\n\\]\nwhich yields the solutions \\( marzipans=1 \\) and, if \\( pinecones^{2} \\neq -1 \\), also \\( marzipans=\\left(pinecones^{2}-1\\right) /\\left(pinecones^{2}+1\\right) \\). (If \\( pinecones^{2}=-1 \\), then \\( 1-pinecones^{2}=-2 \\neq 0 \\), and the second factor gives no solution.) Using \\( raincloud=pinecones(marzipans-1) \\), we find that these give \\( (1,0) \\) and \\( \\left(\\frac{pinecones^{2}-1}{pinecones^{2}+1}, \\frac{-2 pinecones}{pinecones^{2}+1}\\right) \\), which, as we verify, do satisfy \\( marzipans^{2}+raincloud^{2}=1 \\) and \\( raincloud=pinecones(marzipans-1) \\). We finish by substituting \\( pinecones=-quagmire \\).\n\nRemark. Let \\( sapphire \\) be any nondegenerate conic over \\( sandstone \\) with an \\( sandstone \\)-rational point \\( wildfire \\) : this means that \\( sapphire \\) is given by a polynomial \\( compasses(marzipans, raincloud) \\in sandstone[marzipans, raincloud] \\) of total degree 2 that does not factor into linear polynomials over any field extension of \\( sandstone \\), and \\( wildfire=(gravestone, honeycomb) \\in sandstone^{2} \\) is a point such that \\( compasses(gravestone, honeycomb)=0 \\). The same method (of drawing all lines through \\( wildfire \\) with slope in \\( sandstone \\), and seeing where they intersect \\( compasses(marzipans, raincloud)=0 \\) other than at \\( wildfire \\) ) lets one parameterize the set of \\( sandstone \\)-rational points of \\( sapphire \\) in terms of a single parameter \\( pinecones \\). In the language of algebraic geometry, one says that any conic over \\( moonlight \\) with a \\( moonlight \\)-rational point is birationally equivalent to the line [Shaf, p. 11]. The same method works on certain equations of higher degree [NZM, Section 5.6].\n\nThe parameterization in the preceding paragraph is the reason why indefinite integrals of rational functions in \\( drumstick \\) and \\( \\sqrt{afterglow(drumstick)} \\) for a single quadratic polynomial \\( afterglow(drumstick) \\) can be expressed in terms of elementary functions [Shaf, p. 7]. It also explains why \\( \\sin labyrinth \\) and \\( \\cos labyrinth \\) can be expressed as rational functions of a single function:\n\\[\n(\\cos labyrinth, \\sin labyrinth)=\\left(\\frac{1-drumstick^{2}}{1+drumstick^{2}}, \\frac{2 drumstick}{1+drumstick^{2}}\\right) \\quad \\text { where } drumstick=\\tan (labyrinth / 2)\n\\]\nThis, in turn, explains why rational functions in \\( \\sin labyrinth \\) and \\( \\cos labyrinth \\), such as\n\\[\n\\frac{\\sin ^{3} labyrinth-7 \\sin labyrinth \\cos labyrinth}{1+\\cos ^{3} labyrinth}\n\\]\nhave elementary antiderivatives.\n\nRemark. The parameterization of solutions to \\( marzipans^{2}+raincloud^{2}=1 \\) over \\( \\mathbb{Q} \\) is closely linked to the parameterization of primitive Pythagorean triples, i.e., positive integer solutions to \\( gravestone^{2}+honeycomb^{2}=peppercorn^{2} \\) with \\( \\operatorname{gcd}(gravestone, honeycomb, peppercorn)=1 \\). In any primitive Pythagorean triple, exactly one of \\( gravestone \\) and \\( honeycomb \\) is even; the set of primitive Pythagorean triples \\( (gravestone, honeycomb, peppercorn) \\) with \\( honeycomb \\) even equals the set of triples \\( \\left(shipwreck^{2}-snowflake^{2}, 2 shipwreck snowflake, shipwreck^{2}+snowflake^{2}\\right) \\) with \\( shipwreck, snowflake \\) ranging over positive integers of opposite parity satisfying \\( shipwreck>snowflake \\) and \\( \\operatorname{gcd}(shipwreck, snowflake)=1 \\). A more number-theoretic (less geometric) approach to this classification is given in [NZM, Section 5.3]."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "x_r": "upsidedown",
        "y": "horizontalaxis",
        "y_r": "sidewaysup",
        "r": "stillness",
        "s": "levelness",
        "t": "steadiness",
        "a": "conclusion",
        "b": "endingpart",
        "c": "shortside",
        "m": "minimumval",
        "n": "negativeval",
        "p": "constantval",
        "\\\\theta": "zeroradian",
        "L_s": "straightcurve",
        "F": "voidspace",
        "k": "infinitefld",
        "C": "linearity",
        "P": "segmentpt",
        "f": "constantfn"
      },
      "question": "Let $voidspace$ be a field in which $1+1 \\neq 0$. Show that the set of\nsolutions to the equation $verticalaxis^2+horizontalaxis^2=1$ with $verticalaxis$ and $horizontalaxis$ in $voidspace$ is given\nby $(verticalaxis,horizontalaxis)=(1,0)$ and\n\\[\n(verticalaxis,horizontalaxis) = \\left( \\frac{stillness^2-1}{stillness^2+1}, \\frac{2stillness}{stillness^2+1} \\right)\n\\]\nwhere $stillness$ runs through the elements of $voidspace$ such that $stillness^2\\neq -1$.",
      "solution": "Solution 1. For \\( stillness^{2} \\neq-1 \\), let \\( \\left(upsidedown, sidewaysup\\right)=\\left(\\frac{stillness^{2}-1}{stillness^{2}+1}, \\frac{2 stillness}{stillness^{2}+1}\\right) \\). Clearly \\( (1,0) \\) and \\( \\left(upsidedown, sidewaysup\\right) \\) for \\( stillness^{2} \\neq-1 \\) are solutions.\n\nConversely suppose \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\). If \\( verticalaxis=1 \\), then \\( horizontalaxis=0 \\) and we have \\( (1,0) \\). Otherwise define \\( stillness=horizontalaxis /(1-verticalaxis) \\). (The reason for this is that \\( sidewaysup /\\left(1-upsidedown\\right)=stillness \\).) Then \\( 1-verticalaxis^{2}=horizontalaxis^{2}=stillness^{2}(1-verticalaxis)^{2} \\), but \\( verticalaxis \\neq 1 \\), so we may divide by \\( 1-verticalaxis \\) to obtain \\( 1+verticalaxis=stillness^{2}(1-verticalaxis) \\), and \\( \\left(stillness^{2}+1\\right) verticalaxis=\\left(stillness^{2}-1\\right) \\). If \\( stillness^{2}=-1 \\), then this says \\( 0=-2 \\), contradicting \\( 1+1 \\neq 0 \\). Thus \\( stillness^{2} \\neq-1, verticalaxis=\\frac{stillness^{2}-1}{stillness^{2}+1}=upsidedown \\), and \\( horizontalaxis=stillness(1-verticalaxis)=\\frac{2 stillness}{stillness^{2}+1}=sidewaysup \\). Hence every solution to \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\) not equal to \\( (1,0) \\) is of the form \\( \\left(upsidedown, sidewaysup\\right) \\) for some \\( stillness \\in voidspace \\) with \\( stillness^{2} \\neq-1 \\).\n\nRemark. If instead \\( voidspace \\) is a field in which \\( 1+1=0 \\) (i.e., the characteristic of \\( voidspace \\) is 2), then \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\) is equivalent to \\( (verticalaxis+horizontalaxis+1)^{2}=0 \\), and the set of solutions is \\( \\{(steadiness, steadiness+1): steadiness \\in voidspace\\} \\).\n\nSolution 2. Essentially the same solution can be motivated by geometry. The only solution to \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\) with \\( verticalaxis=1 \\) is \\( (1,0) \\). For each \\( (verticalaxis, horizontalaxis) \\in voidspace^{2} \\) satisfying \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\) with \\( verticalaxis \\neq 1 \\), the line through \\( (verticalaxis, horizontalaxis) \\) and \\( (1,0) \\) has slope \\( horizontalaxis /(verticalaxis-1) \\) in \\( voidspace \\). Hence we can find all solutions to \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\) with \\( verticalaxis \\neq 1 \\) by intersecting each nonvertical line through \\( (1,0) \\) with the circle. (See Figure 4.)\n\nIf \\( straightcurve \\) is the line through \\( (1,0) \\) with slope \\( levelness \\in voidspace \\), its intersection with \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\) can be computed by substituting \\( horizontalaxis=levelness( verticalaxis-1) \\) into \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\), and solving the resulting equation\n\\[\nverticalaxis^{2}+levelness^{2}(verticalaxis-1)^{2}=1\n\\]\n\nThis equation is guaranteed to have the solution \\( verticalaxis=1 \\), because \\( (1,0) \\) is in the intersection. Therefore we obtain a factorization,\n\\[\n(verticalaxis-1)\\left(\\left(levelness^{2}+1\\right) verticalaxis+\\left(1-levelness^{2}\\right)\\right)=0\n\\]\nwhich yields the solutions \\( verticalaxis=1 \\) and, if \\( levelness^{2} \\neq-1 \\), also \\( verticalaxis=\\left(levelness^{2}-1\\right) /\\left(levelness^{2}+1\\right) \\). (If \\( levelness^{2}=-1 \\), then \\( 1-levelness^{2}=-2 \\neq 0 \\), and the second factor gives no solution.) Using \\( horizontalaxis=levelness(verticalaxis-1) \\), we find that these give \\( (1,0) \\) and \\( \\left(\\frac{levelness^{2}-1}{levelness^{2}+1}, \\frac{-2 levelness}{levelness^{2}+1}\\right) \\), which, as we verify, do satisfy \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\) and \\( horizontalaxis=levelness(verticalaxis-1) \\). We finish by substituting \\( levelness=-stillness \\).\n\nRemark. Let \\( linearity \\) be any nondegenerate conic over \\( voidspace \\) with an \\( voidspace \\)-rational point \\( segmentpt \\) : this means that \\( linearity \\) is given by a polynomial \\( constantfn(verticalaxis, horizontalaxis) \\in voidspace[verticalaxis, horizontalaxis] \\) of total degree 2 that does not factor into linear polynomials over any field extension of \\( voidspace \\), and \\( segmentpt=(conclusion, endingpart) \\in voidspace^{2} \\) is a point such that \\( constantfn(conclusion, endingpart)=0 \\). The same method (of drawing all lines through \\( segmentpt \\) with slope in \\( voidspace \\), and seeing where they intersect \\( constantfn(verticalaxis, horizontalaxis)=0 \\) other than at \\( segmentpt \\) ) lets one parameterize the set of \\( voidspace \\)-rational points of \\( linearity \\) in terms of a single parameter \\( levelness \\). In the language of algebraic geometry, one says that any conic over \\( infinitefld \\) with a \\( infinitefld \\)-rational point is birationally equivalent to the line [Shaf, p. 11]. The same method works on certain equations of higher degree [NZM, Section 5.6].\n\nThe parameterization in the preceding paragraph is the reason why indefinite integrals of rational functions in \\( steadiness \\) and \\( \\sqrt{constantval(steadiness)} \\) for a single quadratic polynomial \\( constantval(steadiness) \\) can be expressed in terms of elementary functions [Shaf, p. 7]. It also explains why \\( \\sin zeroradian \\) and \\( \\cos zeroradian \\) can be expressed as rational functions of a single function:\n\\[\n(\\cos zeroradian, \\sin zeroradian)=\\left(\\frac{1-steadiness^{2}}{1+steadiness^{2}}, \\frac{2 steadiness}{1+steadiness^{2}}\\right) \\quad \\text { where } steadiness=\\tan (zeroradian / 2)\n\\]\n\nThis, in turn, explains why rational functions in \\( \\sin zeroradian \\) and \\( \\cos zeroradian \\), such as\n\\[\n\\frac{\\sin ^{3} zeroradian-7 \\sin zeroradian \\cos zeroradian}{1+\\cos ^{3} zeroradian}\n\\]\nhave elementary antiderivatives.\nRemark. The parameterization of solutions to \\( verticalaxis^{2}+horizontalaxis^{2}=1 \\) over \\( \\mathbb{Q} \\) is closely linked to the parameterization of primitive Pythagorean triples, i.e., positive integer solutions to \\( conclusion^{2}+endingpart^{2}=shortside^{2} \\) with \\( \\operatorname{gcd}(conclusion, endingpart, shortside)=1 \\). In any primitive Pythagorean triple, exactly one of \\( conclusion \\) and \\( endingpart \\) is even; the set of primitive Pythagorean triples \\( (conclusion, endingpart, shortside) \\) with \\( endingpart \\) even equals the set of triples \\( \\left(minimumval^{2}-negativeval^{2}, 2 minimumval negativeval, minimumval^{2}+negativeval^{2}\\right) \\) with \\( minimumval, negativeval \\) ranging over positive integers of opposite parity satisfying \\( minimumval>negativeval \\) and \\( \\operatorname{gcd}(minimumval, negativeval)=1 \\). A more number-theoretic (less geometric) approach to this classification is given in [NZM, Section 5.3]."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "x_r": "ghsdblrf",
        "y": "hjgrksla",
        "y_r": "vsplckwj",
        "r": "zmqnfkti",
        "s": "pdjrkmhb",
        "t": "nbfwyczp",
        "a": "frlndqsj",
        "b": "vkrtwmhg",
        "c": "mgdwfzrq",
        "m": "tsrvaxlu",
        "n": "hckmqswl",
        "p": "rqdwkzbv",
        "\\theta": "lksjhfmx",
        "L_s": "qmcxprdv",
        "F": "zvfgmlth",
        "k": "sdbnqjrk",
        "C": "bwmqzdft",
        "P": "djrtnkfg",
        "f": "wckzpdhr"
      },
      "question": "Let $zvfgmlth$ be a field in which $1+1 \\neq 0$. Show that the set of\nsolutions to the equation $qzxwvtnp^2+hjgrksla^2=1$ with $qzxwvtnp$ and $hjgrksla$ in $zvfgmlth$ is given\nby $(qzxwvtnp,hjgrksla)=(1,0)$ and\n\\[\n(qzxwvtnp,hjgrksla) = \\left( \\frac{zmqnfkti^2-1}{zmqnfkti^2+1}, \\frac{2zmqnfkti}{zmqnfkti^2+1} \\right)\n\\]\nwhere $zmqnfkti$ runs through the elements of $zvfgmlth$ such that $zmqnfkti^2\\neq -1$.",
      "solution": "Solution 1. For $zmqnfkti^{2} \\neq-1$, let $\\left(qzxwvtnp_{zmqnfkti}, hjgrksla_{zmqnfkti}\\right)=\\left(\\frac{zmqnfkti^{2}-1}{zmqnfkti^{2}+1}, \\frac{2 zmqnfkti}{zmqnfkti^{2}+1}\\right)$. Clearly $(1,0)$ and $\\left(qzxwvtnp_{zmqnfkti}, hjgrksla_{zmqnfkti}\\right)$ for $zmqnfkti^{2} \\neq-1$ are solutions.\n\nConversely suppose $qzxwvtnp^{2}+hjgrksla^{2}=1$. If $qzxwvtnp=1$, then $hjgrksla=0$ and we have $(1,0)$. Otherwise define $zmqnfkti=hjgrksla /(1-qzxwvtnp)$. (The reason for this is that $hjgrksla_{zmqnfkti} /\\left(1-qzxwvtnp_{zmqnfkti}\\right)=zmqnfkti$.) Then $1-qzxwvtnp^{2}=hjgrksla^{2}=zmqnfkti^{2}(1-qzxwvtnp)^{2}$, but $qzxwvtnp \\neq 1$, so we may divide by $1-qzxwvtnp$ to obtain $1+qzxwvtnp=zmqnfkti^{2}(1-qzxwvtnp)$, and $(zmqnfkti^{2}+1) qzxwvtnp=(zmqnfkti^{2}-1)$. If $zmqnfkti^{2}=-1$, then this says $0=-2$, contradicting $1+1 \\neq 0$. Thus $zmqnfkti^{2} \\neq-1$, $qzxwvtnp=\\frac{zmqnfkti^{2}-1}{zmqnfkti^{2}+1}=qzxwvtnp_{zmqnfkti}$, and $hjgrksla=zmqnfkti(1-qzxwvtnp)=\\frac{2 zmqnfkti}{zmqnfkti^{2}+1}=hjgrksla_{zmqnfkti}$. Hence every solution to $qzxwvtnp^{2}+hjgrksla^{2}=1$ not equal to $(1,0)$ is of the form $\\left(qzxwvtnp_{zmqnfkti}, hjgrksla_{zmqnfkti}\\right)$ for some $zmqnfkti \\in zvfgmlth$ with $zmqnfkti^{2} \\neq-1$.\n\nRemark. If instead $zvfgmlth$ is a field in which $1+1=0$ (i.e., the characteristic of $zvfgmlth$ is 2), then $qzxwvtnp^{2}+hjgrksla^{2}=1$ is equivalent to $(qzxwvtnp+hjgrksla+1)^{2}=0$, and the set of solutions is $\\{(nbfwyczp, nbfwyczp+1): nbfwyczp \\in zvfgmlth\\}$.\n\nSolution 2. Essentially the same solution can be motivated by geometry. The only solution to $qzxwvtnp^{2}+hjgrksla^{2}=1$ with $qzxwvtnp=1$ is $(1,0)$. For each $(qzxwvtnp, hjgrksla) \\in zvfgmlth^{2}$ satisfying $qzxwvtnp^{2}+hjgrksla^{2}=1$ with $qzxwvtnp \\neq 1$, the line through $(qzxwvtnp, hjgrksla)$ and $(1,0)$ has slope $hjgrksla /(qzxwvtnp-1)$ in $zvfgmlth$. Hence we can find all solutions to $qzxwvtnp^{2}+hjgrksla^{2}=1$ with $qzxwvtnp \\neq 1$ by intersecting each nonvertical line through $(1,0)$ with the circle. (See Figure 4.)\n\nIf $qmcxprdv$ is the line through $(1,0)$ with slope $pdjrkmhb \\in zvfgmlth$, its intersection with $qzxwvtnp^{2}+hjgrksla^{2}=1$ can be computed by substituting $hjgrksla=pdjrkmhb(qzxwvtnp-1)$ into $qzxwvtnp^{2}+hjgrksla^{2}=1$, and solving the resulting equation\n\\[\nqzxwvtnp^{2}+pdjrkmhb^{2}(qzxwvtnp-1)^{2}=1\n\\]\n\nThis equation is guaranteed to have the solution $qzxwvtnp=1$, because $(1,0)$ is in the intersection. Therefore we obtain a factorization,\n\\[\n(qzxwvtnp-1)\\left((pdjrkmhb^{2}+1) qzxwvtnp+(1-pdjrkmhb^{2})\\right)=0\n\\]\nwhich yields the solutions $qzxwvtnp=1$ and, if $pdjrkmhb^{2} \\neq-1$, also $qzxwvtnp=(pdjrkmhb^{2}-1)/(pdjrkmhb^{2}+1)$. (If $pdjrkmhb^{2}=-1$, then $1-pdjrkmhb^{2}=-2 \\neq 0$, and the second factor gives no solution.) Using $hjgrksla=pdjrkmhb(qzxwvtnp-1)$, we find that these give $(1,0)$ and $\\left(\\frac{pdjrkmhb^{2}-1}{pdjrkmhb^{2}+1}, \\frac{-2 pdjrkmhb}{pdjrkmhb^{2}+1}\\right)$, which, as we verify, do satisfy $qzxwvtnp^{2}+hjgrksla^{2}=1$ and $hjgrksla=pdjrkmhb(qzxwvtnp-1)$. We finish by substituting $pdjrkmhb=-zmqnfkti$.\n\nRemark. Let $bwmqzdft$ be any nondegenerate conic over $zvfgmlth$ with an $zvfgmlth$-rational point $djrtnkfg$ : this means that $bwmqzdft$ is given by a polynomial $wckzpdhr(qzxwvtnp, hjgrksla) \\in zvfgmlth[qzxwvtnp, hjgrksla]$ of total degree 2 that does not factor into linear polynomials over any field extension of $zvfgmlth$, and $djrtnkfg=(frlndqsj, vkrtwmhg) \\in zvfgmlth^{2}$ is a point such that $wckzpdhr(frlndqsj, vkrtwmhg)=0$. The same method (of drawing all lines through $djrtnkfg$ with slope in $zvfgmlth$, and seeing where they intersect $wckzpdhr(qzxwvtnp, hjgrksla)=0$ other than at $djrtnkfg$ ) lets one parameterize the set of $zvfgmlth$-rational points of $bwmqzdft$ in terms of a single parameter $pdjrkmhb$. In the language of algebraic geometry, one says that any conic over $sdbnqjrk$ with a $sdbnqjrk$-rational point is birationally equivalent to the line [Shaf, p. 11]. The same method works on certain equations of higher degree [NZM, Section 5.6].\n\nThe parameterization in the preceding paragraph is the reason why indefinite integrals of rational functions in $nbfwyczp$ and $\\sqrt{rqdwkzbv(nbfwyczp)}$ for a single quadratic polynomial $rqdwkzbv(nbfwyczp)$ can be expressed in terms of elementary functions [Shaf, p. 7]. It also explains why $\\sin lksjhfmx$ and $\\cos lksjhfmx$ can be expressed as rational functions of a single function:\n\\[\n(\\cos lksjhfmx, \\sin lksjhfmx)=\\left(\\frac{1-nbfwyczp^{2}}{1+nbfwyczp^{2}}, \\frac{2 nbfwyczp}{1+nbfwyczp^{2}}\\right) \\quad \\text { where } nbfwyczp=\\tan (lksjhfmx / 2)\n\\]\n\nThis, in turn, explains why rational functions in $\\sin lksjhfmx$ and $\\cos lksjhfmx$, such as\n\\[\n\\frac{\\sin ^{3} lksjhfmx-7 \\sin lksjhfmx \\cos lksjhfmx}{1+\\cos ^{3} lksjhfmx}\n\\]\nhave elementary antiderivatives.\nRemark. The parameterization of solutions to $qzxwvtnp^{2}+hjgrksla^{2}=1$ over $\\mathbb{Q}$ is closely linked to the parameterization of primitive Pythagorean triples, i.e., positive integer solutions to $frlndqsj^{2}+vkrtwmhg^{2}=mgdwfzrq^{2}$ with $\\operatorname{gcd}(frlndqsj, vkrtwmhg, mgdwfzrq)=1$. In any primitive Pythagorean triple, exactly one of $frlndqsj$ and $vkrtwmhg$ is even; the set of primitive Pythagorean triples $(frlndqsj, vkrtwmhg, mgdwfzrq)$ with $vkrtwmhg$ even equals the set of triples $\\left(tsrvaxlu^{2}-hckmqswl^{2}, 2 tsrvaxlu hckmqswl, tsrvaxlu^{2}+hckmqswl^{2}\\right)$ with $tsrvaxlu, hckmqswl$ ranging over positive integers of opposite parity satisfying $tsrvaxlu>hckmqswl$ and $\\operatorname{gcd}(tsrvaxlu, hckmqswl)=1$. A more number-theoretic (less geometric) approach to this classification is given in [NZM, Section 5.3]."
    },
    "kernel_variant": {
      "question": "Let $F$ be a field with $1+1\\neq 0$.  Determine all pairs $(x,y)\\in F^{2}$ satisfying\n\\[\n x^{2}+y^{2}=4.\n\\]",
      "solution": "First observe that (2,0) is a solution of x²+y²=4.\n\nSuppose now that (x,y) is any other solution, so x≠2.  Introduce the slope parameter\n  r:=y/(2−x)∈F,\nso that y=r(2−x).  Substituting this into x²+y²=4 gives\n  x²+r²(2−x)²=4.\nSince (2,0) satisfies the same equation, the polynomial\n  x²+r²(2−x)²−4\nvanishes at x=2 and hence factors as\n  (x−2)((r²+1)x+2(1−r²))=0.\nBecause x≠2, the second factor must be zero, so\n  (r²+1)x+2(1−r²)=0.\nIf r²=−1, this would force 4=0 and hence 1+1=0, contradicting the hypothesis; hence r²≠−1 and we may divide by r²+1 to obtain\n  x=2(r²−1)/(r²+1),\nand then\n  y=r(2−x)=4r/(r²+1).\n\nConversely, a direct calculation shows\n  (2(r²−1)/(r²+1))²+(4r/(r²+1))²\n=4(r⁴+2r²+1)/(r²+1)²=4.\nTherefore the complete set of solutions in F² to x²+y²=4 is the point (2,0) together with the one–parameter family\n  (x,y)= (2(r²−1)/(r²+1), 4r/(r²+1)),\nwhere r runs through F with r²≠−1.",
      "_meta": {
        "core_steps": [
          "Note the obvious solution (1,0).",
          "For any other solution define r = y/(1−x) (slope of the line through the fixed point).",
          "Substitute y = r(1−x) into x^2 + y^2 = 1 and factor out (1−x) to obtain (r^2+1)x = r^2−1.",
          "Rule out r^2 = −1 using 1+1 ≠ 0 and solve to get x = (r^2−1)/(r^2+1), y = 2r/(r^2+1).",
          "Collect the x = 1 case with the parametrised family to list all solutions."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Chosen F-rational point through which lines are drawn in the parametrisation argument.",
            "original": "(1,0)"
          },
          "slot2": {
            "description": "Value on the right-hand side of the quadratic form; any non-zero element would work after scaling.",
            "original": "1 in the equation x^2 + y^2 = 1"
          },
          "slot3": {
            "description": "Specific slope parameter chosen; other algebraically convenient re-definitions (e.g. −y/(1−x), y/(x−1)) also work.",
            "original": "r = y/(1−x)"
          },
          "slot4": {
            "description": "Factor that appears in the y-coordinate of the parametrisation as a consequence of the chosen base point and slope definition.",
            "original": "2 in y = 2r/(r^2+1)"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}