{
  "index": "1954-A-7",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "7. Prove that there are no integers \\( x \\) and \\( y \\) for which\n\\[\nx^{2}+3 x y-2 y^{2}=122\n\\]",
  "solution": "Solution. Multiply (1) by 4 and complete the square to obtain\n\\[\n(2 x+3 y)^{2}-17 y^{2}=488\n\\]\n\nIf (1) has a solution in integers, then so does (2) and hence so does the congruence\n\\[\nu^{2} \\equiv 488(\\bmod 17) .\n\\]\n\nBut (3) has no solutions because 488 is not a quadratic residue of 17 . The most direct way to see this is to check the possibilities \\( u=0,1, \\ldots, 8 \\). (We need not worry about \\( u=9, \\ldots, 16 \\) because \\( u^{2} \\equiv(17-u)^{2} \\).) Alternatively we can calculate using the quadratic reciprocity theorem and the Legendre symbol. We have\n\\[\n(488 / 17)=(12 / 17)=(2 / 17)^{2}(3 / 17)=(3 / 17)=(17 / 3)=(2 / 3)=-1 ;\n\\]\nhence 488 is not a quadratic residue of 17 .\nRemark. We could equally well argue that a solution of (1) and hence of (2) in integers would imply that 17 is a quadratic residue of 61 , since 61 divides 488 . But \\( (17 / 61)=(61 / 17)=(10 / 17)=(2 / 17)(5 / 17)= \\) \\( 1 \\cdot(17 / 5)=(2 / 5)=-1 \\).\n\nFor the quadratic reciprocity theorem and the Legendre symbol see, for example, W. J. LeVeque, Topics in Number Theory, vol. 1, AddisonWesley, Reading, Mass., 1956, page 66 ff.",
  "vars": [
    "x",
    "y",
    "u"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "integerx",
        "y": "integery",
        "u": "modularu"
      },
      "question": "7. Prove that there are no integers \\( integerx \\) and \\( integery \\) for which\n\\[\nintegerx^{2}+3 integerx integery-2 integery^{2}=122\n\\]",
      "solution": "Solution. Multiply (1) by 4 and complete the square to obtain\n\\[\n(2 integerx+3 integery)^{2}-17 integery^{2}=488\n\\]\n\nIf (1) has a solution in integers, then so does (2) and hence so does the congruence\n\\[\nmodularu^{2} \\equiv 488(\\bmod 17) .\n\\]\n\nBut (3) has no solutions because 488 is not a quadratic residue of 17 . The most direct way to see this is to check the possibilities \\( modularu=0,1, \\ldots, 8 \\). (We need not worry about \\( modularu=9, \\ldots, 16 \\) because \\( modularu^{2} \\equiv(17-modularu)^{2} \\).) Alternatively we can calculate using the quadratic reciprocity theorem and the Legendre symbol. We have\n\\[\n(488 / 17)=(12 / 17)=(2 / 17)^{2}(3 / 17)=(3 / 17)=(17 / 3)=(2 / 3)=-1 ;\n\\]\nhence 488 is not a quadratic residue of 17 .\nRemark. We could equally well argue that a solution of (1) and hence of (2) in integers would imply that 17 is a quadratic residue of 61 , since 61 divides 488 . But \\( (17 / 61)=(61 / 17)=(10 / 17)=(2 / 17)(5 / 17)= \\) \\( 1 \\cdot(17 / 5)=(2 / 5)=-1 \\).\n\nFor the quadratic reciprocity theorem and the Legendre symbol see, for example, W. J. LeVeque, Topics in Number Theory, vol. 1, AddisonWesley, Reading, Mass., 1956, page 66 ff."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "y": "chocolate",
        "u": "squirrelly"
      },
      "question": "7. Prove that there are no integers \\( pineapple \\) and \\( chocolate \\) for which\n\\[\npineapple^{2}+3 pineapple chocolate-2 chocolate^{2}=122\n\\]",
      "solution": "Solution. Multiply (1) by 4 and complete the square to obtain\n\\[\n(2 pineapple+3 chocolate)^{2}-17 chocolate^{2}=488\n\\]\n\nIf (1) has a solution in integers, then so does (2) and hence so does the congruence\n\\[\nsquirrelly^{2} \\equiv 488(\\bmod 17) .\n\\]\n\nBut (3) has no solutions because 488 is not a quadratic residue of 17 . The most direct way to see this is to check the possibilities \\( squirrelly=0,1, \\ldots, 8 \\). (We need not worry about \\( squirrelly=9, \\ldots, 16 \\) because \\( squirrelly^{2} \\equiv(17-squirrelly)^{2} \\).) Alternatively we can calculate using the quadratic reciprocity theorem and the Legendre symbol. We have\n\\[\n(488 / 17)=(12 / 17)=(2 / 17)^{2}(3 / 17)=(3 / 17)=(17 / 3)=(2 / 3)=-1 ;\n\\]\nhence 488 is not a quadratic residue of 17 .\nRemark. We could equally well argue that a solution of (1) and hence of (2) in integers would imply that 17 is a quadratic residue of 61 , since 61 divides 488 . But \\( (17 / 61)=(61 / 17)=(10 / 17)=(2 / 17)(5 / 17)= \\) \\( 1 \\cdot(17 / 5)=(2 / 5)=-1 \\).\n\nFor the quadratic reciprocity theorem and the Legendre symbol see, for example, W. J. LeVeque, Topics in Number Theory, vol. 1, AddisonWesley, Reading, Mass., 1956, page 66 ff."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "u": "constantvalue"
      },
      "question": "7. Prove that there are no integers \\( verticalaxis \\) and \\( horizontalaxis \\) for which\n\\[\nverticalaxis^{2}+3 verticalaxis horizontalaxis-2 horizontalaxis^{2}=122\n\\]",
      "solution": "Solution. Multiply (1) by 4 and complete the square to obtain\n\\[\n(2 verticalaxis+3 horizontalaxis)^{2}-17 horizontalaxis^{2}=488\n\\]\n\nIf (1) has a solution in integers, then so does (2) and hence so does the congruence\n\\[\nconstantvalue^{2} \\equiv 488(\\bmod 17) .\n\\]\n\nBut (3) has no solutions because 488 is not a quadratic residue of 17 . The most direct way to see this is to check the possibilities \\( constantvalue=0,1, \\ldots, 8 \\). (We need not worry about \\( constantvalue=9, \\ldots, 16 \\) because \\( constantvalue^{2} \\equiv(17-constantvalue)^{2} \\).) Alternatively we can calculate using the quadratic reciprocity theorem and the Legendre symbol. We have\n\\[\n(488 / 17)=(12 / 17)=(2 / 17)^{2}(3 / 17)=(3 / 17)=(17 / 3)=(2 / 3)=-1 ;\n\\]\nhence 488 is not a quadratic residue of 17 .\nRemark. We could equally well argue that a solution of (1) and hence of (2) in integers would imply that 17 is a quadratic residue of 61 , since 61 divides 488 . But \\( (17 / 61)=(61 / 17)=(10 / 17)=(2 / 17)(5 / 17)= \\) \\( 1 \\cdot(17 / 5)=(2 / 5)=-1 \\).\n\nFor the quadratic reciprocity theorem and the Legendre symbol see, for example, W. J. LeVeque, Topics in Number Theory, vol. 1, AddisonWesley, Reading, Mass., 1956, page 66 ff."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "u": "bnmxvzqe"
      },
      "question": "7. Prove that there are no integers \\( qzxwvtnp \\) and \\( hjgrksla \\) for which\n\\[\nqzxwvtnp^{2}+3 qzxwvtnp hjgrksla-2 hjgrksla^{2}=122\n\\]",
      "solution": "Solution. Multiply (1) by 4 and complete the square to obtain\n\\[\n(2 qzxwvtnp+3 hjgrksla)^{2}-17 hjgrksla^{2}=488\n\\]\n\nIf (1) has a solution in integers, then so does (2) and hence so does the congruence\n\\[\nbnmxvzqe^{2} \\equiv 488(\\bmod 17) .\n\\]\n\nBut (3) has no solutions because 488 is not a quadratic residue of 17 . The most direct way to see this is to check the possibilities \\( bnmxvzqe=0,1, \\ldots, 8 \\). (We need not worry about \\( bnmxvzqe=9, \\ldots, 16 \\) because \\( bnmxvzqe^{2} \\equiv(17-bnmxvzqe)^{2} \\).) Alternatively we can calculate using the quadratic reciprocity theorem and the Legendre symbol. We have\n\\[\n(488 / 17)=(12 / 17)=(2 / 17)^{2}(3 / 17)=(3 / 17)=(17 / 3)=(2 / 3)=-1 ;\n\\]\nhence 488 is not a quadratic residue of 17 .\nRemark. We could equally well argue that a solution of (1) and hence of (2) in integers would imply that 17 is a quadratic residue of 61 , since 61 divides 488 . But \\( (17 / 61)=(61 / 17)=(10 / 17)=(2 / 17)(5 / 17)= \\) \\( 1 \\cdot(17 / 5)=(2 / 5)=-1 \\).\n\nFor the quadratic reciprocity theorem and the Legendre symbol see, for example, W. J. LeVeque, Topics in Number Theory, vol. 1, AddisonWesley, Reading, Mass., 1956, page 66 ff."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nf(x,y)=x^{2}+3xy-2y^{2}, \\qquad (x,y\\in\\mathbb Z),\\qquad \n\\Delta(f)=17 .\n\\]\n\nFix a non–zero integer  \n\\[\nN,\\qquad \\gcd(N,17)=1,\\qquad  \nN=\\prod_{i=1}^{s}p_{i}^{\\,e_{i}}\\;(e_{i}\\ge 1),\n\\]\nwhere the $p_{i}$ are pairwise distinct odd primes.\n\nFor every odd prime $p\\neq 17$ put  \n\\[\n\\left(\\dfrac{17}{p}\\right)=\n\\begin{cases}\n\\;1 &\\text{if }17\\text{ is a quadratic residue modulo }p,\\\\[6pt]\n-1 &\\text{otherwise}.\n\\end{cases}\n\\]\n\nDefine the two index sets  \n\\[\nI=\\Bigl\\{\\,i\\mid \\bigl(\\tfrac{17}{p_{i}}\\bigr)=-1\\Bigr\\},\n\\qquad\nI^{c}=\\{1,\\dots ,s\\}\\setminus I .\n\\]\n\n1. (Global $\\Longrightarrow$ local parity obstruction)  \n   Show that if  \n   \\[\n   f(x,y)=N\n   \\]\n   has an integral solution, then  \n   \\[\n   e_{i}\\equiv 0\\pmod{2}\\qquad\\text{for every }i\\in I .\n   \\tag{P}\n   \\]\n\n2. (Local–global principle for the form $f$)  \n   Prove the converse of (P).  \n   Treat the two cases $N>0$ and $N<0$ separately, and make essential use of the following facts about the quadratic field  \n   \\[\n   K=\\mathbb Q(\\sqrt{17}):\n   \\qquad h^{+}(K)=1,\\qquad\n   \\varepsilon=33+8\\sqrt{17}\\text{ is a totally positive unit of norm }1,\n   \\qquad\n   \\tau =4+\\sqrt{17}\\text{ has norm }-1 .\n   \\]\n   Deduce  \n   \\[\n   f(x,y)=N\\text{ is solvable in }\\mathbb Z^{2}\n   \\;\\Longleftrightarrow\\;\n   e_{i}\\equiv 0\\pmod 2\\text{ for every }i\\in I .\n   \\tag{$\\diamondsuit$}\n   \\]\n\n3. (Purely local reformulation)  \n   Prove that $(\\diamondsuit)$ is equivalent to  \n\n   “for every prime power $p^{k}$ the congruence  \n   $f(x,y)\\equiv N \\pmod{p^{k}}$ is solvable.”\n\n   Hence verify that $f$ satisfies the Hasse–Minkowski principle.\n\n4. Apply $(\\diamondsuit)$ to decide whether the equation  \n   \\[\n   x^{2}+3xy-2y^{2}=122\n   \\]\n   possesses integral solutions.\n\n5. (“One–bit” obstruction)  \n   Show that the weaker parity test\n   \\[\n   \\sum_{i\\in I} e_{i}\\text{ odd}\\;\\Longrightarrow\\;\n   f(x,y)=N\\text{ has no integral solution},\n   \\]\n   thereby recovering the classical criterion, and explain why it is\n   insufficient in the present context.\n\n\\bigskip",
      "solution": "Throughout we set  \n\\[\nK=\\mathbb Q(\\sqrt{17}),\\qquad \n\\mathcal O_{K}=\\mathbb Z[\\omega],\\; \\omega=\\tfrac{1+\\sqrt{17}}{2}.\n\\]\n\nStep 0.  A norm form equivalent to $f$.  \nDefine  \n\\[\ng(x,y)=x^{2}+xy-4y^{2}.\n\\]\nThe unimodular substitution $(x,y)\\mapsto(x+y,y)$ yields $g(x+y,y)=f(x,y)$, so $f$ represents $N$ over $\\mathbb Z$ iff $g$ does.  \nFurthermore  \n\\[\nN_{K/\\mathbb Q}(x+y\\omega)=(x+y\\omega)(x+y\\bar\\omega)=g(x,y).\n\\tag{0}\n\\]\n\nStep 1.  Necessity of the parity condition (P).  \nAssume $f(x_{0},y_{0})=N$.  Then $N=g(x_{0}+y_{0},y_{0})\n     =N_{K/\\mathbb Q}(\\alpha_{0})$ with $\\alpha_{0}=x_{0}+y_{0}\\omega\\in\\mathcal O_{K}$.  \n\nLet $p$ be inert, i.e.\\ $\\bigl(\\tfrac{17}{p}\\bigr)=-1$.  \nThen $(p)$ remains prime in $\\mathcal O_{K}$, say $(p)=\\mathfrak P=\\bar{\\mathfrak P}$, and for every $\\beta\\in\\mathcal O_{K}$ we have  \n\\[\nv_{p}\\bigl(N_{K/\\mathbb Q}(\\beta)\\bigr)\n      =2\\,v_{\\mathfrak P}(\\beta)\n\\]\nso is even.  Applying this to $\\beta=\\alpha_{0}$ gives  \n$v_{p}(N)=e_{p}$ even, establishing (P).\n\nStep 2.  Sufficiency of (P).  \n\n2.1  Ideal factorisation.  \nWrite  \n\\[\nN=\\Bigl(\\prod_{i\\in I}p_{i}^{e_{i}}\\Bigr)\n   \\Bigl(\\prod_{j\\in I^{c}}p_{j}^{e_{j}}\\Bigr).\n\\]\nFor every $j\\in I^{c}$ choose one of the two prime ideals above $p_{j}$ and\ncall it $\\mathfrak p_{j}$.  \nFor $i\\in I$ the only prime ideal is $\\mathfrak p_{i}=(p_{i})$.  \nBecause $e_{i}$ is even, put $e_{i}=2m_{i}$.  Set  \n\\[\n\\mathfrak a:=\\prod_{i\\in I}\\mathfrak p_{i}^{\\,m_{i}}\n            \\prod_{j\\in I^{c}}\\mathfrak p_{j}^{\\,e_{j}},\n\\qquad\n(N)=\\mathfrak a\\,\\bar{\\mathfrak a}.\n\\tag{1}\n\\]\n\n2.2  Principality in the \\emph{narrow} class group.  \nSince $h^{+}(K)=1$, the ideal $\\mathfrak a$ is generated by a\n\\emph{totally positive} element; choose $\\beta\\in\\mathcal O_{K}$ with  \n\\[\n\\mathfrak a=(\\beta),\\qquad\n\\beta\\succ 0.\n\\]\nTaking norms in (1) gives  \n\\[\nN_{K/\\mathbb Q}(\\beta)=u\\cdot |N|,\\qquad u\\in\\mathcal O_{K}^{\\times},\\quad\nN_{K/\\mathbb Q}(u)=\\pm1 .\n\\]\nBecause $\\beta$ is totally positive, its norm is positive, forcing\n$N_{K/\\mathbb Q}(u)=1$.  Hence  \n\\[\nN_{K/\\mathbb Q}(\\beta)=|N|.\n\\tag{2}\n\\]\n\n2.3  Adjusting the sign.  \nLet $\\tau:=4+\\sqrt{17}$; then $N_{K/\\mathbb Q}(\\tau)=-1$.  Define  \n\\[\n\\alpha=\n\\begin{cases}\n\\beta &\\text{if }N>0,\\\\[6pt]\n\\tau\\beta &\\text{if }N<0.\n\\end{cases}\n\\]\nIn both cases $\\alpha\\in\\mathcal O_{K}$ and  \n\\[\nN_{K/\\mathbb Q}(\\alpha)=N.\n\\tag{3}\n\\]\n\n2.4  Returning to quadratic forms.  \nWrite $\\alpha=x+y\\omega$ with $x,y\\in\\mathbb Z$.  \nBy (0) and (3) we have $g(x,y)=N$, whence  \n\\[\nf(x-y,y)=N,\n\\]\nso $(x-y,y)$ is an integral solution.  \nThus (P) $\\Longrightarrow$ solvability, completing the proof of $(\\diamondsuit)$.\n\nStep 3.  Local solvability and the Hasse–Minkowski principle.  \n\n3.1  $(\\diamondsuit)\\Rightarrow$ local solvability.  \nAn integral solution gives one modulo every $p^{k}$.\n\n3.2  Local solvability $\\Longrightarrow$ (P).  \nAssume that \\emph{for every prime power $p^{k}$} the congruence  \n\\[\nf(x,y)\\equiv N\\pmod{p^{k}}\n\\tag{4}\n\\]\nis solvable.  Fix an inert prime $p$ and pass to the\n$p$–adic completion.  As $f$ is unimodularly equivalent to the norm form\n$g$, (4) yields an element $\\alpha\\in\\mathcal O_{K}\\otimes\\mathbb Z_{p}$\nwith  \n\\[\nN_{K/\\mathbb Q}(\\alpha)=N.\n\\tag{5}\n\\]\n\nLet $v_{p}$ be the $p$–adic valuation on $\\mathbb Q_{p}$ and extend it\nto $K\\otimes\\mathbb Q_{p}$ (there is a \\emph{unique} extension because\n$p$ is unramified in $K$).  For every $\\beta$ in this local field\n\\[\nv_{p}\\bigl(N_{K/\\mathbb Q}(\\beta)\\bigr)=2\\,v_{p}(\\beta).\n\\tag{6}\n\\]\nIndeed, the residue field extension is still degree $2$, hence unramified,\nso the uniformiser of $\\mathcal O_{K}\\otimes\\mathbb Z_{p}$ is a uniformiser\nof $\\mathbb Z_{p}$, and the norm multiplies valuations by $2$.\n\nApplying (6) to $\\alpha$ in (5) we get  \n\\[\nv_{p}(N)=2\\,v_{p}(\\alpha)\\equiv 0\\pmod 2 .\n\\]\nTherefore $e_{p}=v_{p}(N)$ is even for every inert $p$, i.e.\\ (P) holds.\n\nConsequently “complete local solvability'' is equivalent to (P).\nTogether with Step 2 this proves that $f$ satisfies the\nHasse–Minkowski principle.\n\nStep 4.  Application to $N=122=2\\cdot 61$.  \nCompute  \n\\[\n\\left(\\dfrac{17}{2}\\right)=1,\\qquad\n\\left(\\dfrac{17}{61}\\right)=\n\\left(\\dfrac{61}{17}\\right)=\n\\left(\\dfrac{10}{17}\\right)=\n\\left(\\dfrac{2}{17}\\right)\\left(\\dfrac{5}{17}\\right)=1\\cdot(-1)=-1.\n\\]\nHence $61$ is inert and occurs with exponent $1$, violating (P).  \nBy $(\\diamondsuit)$ the equation  \n\\[\nx^{2}+3xy-2y^{2}=122\n\\]\nhas \\emph{no} integer solutions.\n\nStep 5.  The “one–bit’’ obstruction.  \nFrom Step 1, if $f$ represents $N$ then every inert prime occurs with an\n\\emph{even} exponent, so  \n\\[\n\\left(\\dfrac{N}{17}\\right)\n   =\\prod_{i\\in I}\\left(\\dfrac{17}{p_{i}}\\right)^{e_{i}}\n   =(-1)^{\\sum_{i\\in I}e_{i}}=1 .\n\\]\nThus an odd \\emph{total} number of inert prime factors already prevents\nsolvability, reproducing the classical criterion.  However this is\nstrictly weaker than (P): parity has to be checked \\emph{prime by prime}\n(e.g.\\ $N=15=3\\cdot5$ fails $(\\diamondsuit)$ although the total number\nof inert primes is even).\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.469421",
        "was_fixed": false,
        "difficulty_analysis": "•  The original (and kernel) problems required only a single congruence\n   argument modulo one prime; the enhanced variant demands a full local–global\n   characterisation of the values taken by the quadratic form.\n\n•  To prove (*), one must\n   – transform a binary quadratic form into a norm form,  \n   – work in the quadratic integer ring ℤ[√17],  \n   – use the unit group of that ring (via the Pell equation) to lift local\n     information to an integral solution, i.e. employ Dirichlet’s “Pell\n     trick’’/Hasse–Minkowski type reasoning,  \n   – combine these observations with quadratic reciprocity.\n\n•  Part (**) intertwines congruence conditions modulo 4 and 17, demanding\n   simultaneous control of two quadratic symbols and multiplicativity; this\n   yields an infinite family of forbidden integers rather than a single\n   numerical instance.\n\n•  Altogether the enhanced problem obliges the solver to master\n   local solvability, norm forms, the structure of units in real quadratic\n   fields, the multiplicativity of Legendre symbols, and quadratic reciprocity\n   – a substantial escalation in both technical depth and conceptual\n   breadth compared with the single-step residue-check of the kernel task."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nf(x,y)=x^{2}+3xy-2y^{2}, \\qquad (x,y\\in\\mathbb Z),\\qquad \n\\Delta(f)=17 .\n\\]\n\nFix a non–zero integer  \n\\[\nN,\\qquad \\gcd(N,17)=1,\\qquad  \nN=\\prod_{i=1}^{s}p_{i}^{\\,e_{i}}\\;(e_{i}\\ge 1),\n\\]\nwhere the $p_{i}$ are pairwise distinct odd primes.\n\nFor every odd prime $p\\neq 17$ put  \n\\[\n\\left(\\dfrac{17}{p}\\right)=\n\\begin{cases}\n\\;1 &\\text{if }17\\text{ is a quadratic residue modulo }p,\\\\[6pt]\n-1 &\\text{otherwise}.\n\\end{cases}\n\\]\n\nDefine the two index sets  \n\\[\nI=\\Bigl\\{\\,i\\mid \\bigl(\\tfrac{17}{p_{i}}\\bigr)=-1\\Bigr\\},\n\\qquad\nI^{c}=\\{1,\\dots ,s\\}\\setminus I .\n\\]\n\n1. (Global $\\Longrightarrow$ local parity obstruction)  \n   Show that if  \n   \\[\n   f(x,y)=N\n   \\]\n   has an integral solution, then  \n   \\[\n   e_{i}\\equiv 0\\pmod{2}\\qquad\\text{for every }i\\in I .\n   \\tag{P}\n   \\]\n\n2. (Local–global principle for the form $f$)  \n   Prove the converse of (P).  \n   Treat the two cases $N>0$ and $N<0$ separately, and make essential use of the following facts about the quadratic field  \n   \\[\n   K=\\mathbb Q(\\sqrt{17}):\n   \\qquad h^{+}(K)=1,\\qquad\n   \\varepsilon=33+8\\sqrt{17}\\text{ is a totally positive unit of norm }1,\n   \\qquad\n   \\tau =4+\\sqrt{17}\\text{ has norm }-1 .\n   \\]\n   Deduce  \n   \\[\n   f(x,y)=N\\text{ is solvable in }\\mathbb Z^{2}\n   \\;\\Longleftrightarrow\\;\n   e_{i}\\equiv 0\\pmod 2\\text{ for every }i\\in I .\n   \\tag{$\\diamondsuit$}\n   \\]\n\n3. (Purely local reformulation)  \n   Prove that $(\\diamondsuit)$ is equivalent to  \n\n   “for every prime power $p^{k}$ the congruence  \n   $f(x,y)\\equiv N \\pmod{p^{k}}$ is solvable.”\n\n   Hence verify that $f$ satisfies the Hasse–Minkowski principle.\n\n4. Apply $(\\diamondsuit)$ to decide whether the equation  \n   \\[\n   x^{2}+3xy-2y^{2}=122\n   \\]\n   possesses integral solutions.\n\n5. (“One–bit” obstruction)  \n   Show that the weaker parity test\n   \\[\n   \\sum_{i\\in I} e_{i}\\text{ odd}\\;\\Longrightarrow\\;\n   f(x,y)=N\\text{ has no integral solution},\n   \\]\n   thereby recovering the classical criterion, and explain why it is\n   insufficient in the present context.\n\n\\bigskip",
      "solution": "Throughout we set  \n\\[\nK=\\mathbb Q(\\sqrt{17}),\\qquad \n\\mathcal O_{K}=\\mathbb Z[\\omega],\\; \\omega=\\tfrac{1+\\sqrt{17}}{2}.\n\\]\n\nStep 0.  A norm form equivalent to $f$.  \nDefine  \n\\[\ng(x,y)=x^{2}+xy-4y^{2}.\n\\]\nThe unimodular substitution $(x,y)\\mapsto(x+y,y)$ yields $g(x+y,y)=f(x,y)$, so $f$ represents $N$ over $\\mathbb Z$ iff $g$ does.  \nFurthermore  \n\\[\nN_{K/\\mathbb Q}(x+y\\omega)=(x+y\\omega)(x+y\\bar\\omega)=g(x,y).\n\\tag{0}\n\\]\n\nStep 1.  Necessity of the parity condition (P).  \nAssume $f(x_{0},y_{0})=N$.  Then $N=g(x_{0}+y_{0},y_{0})\n     =N_{K/\\mathbb Q}(\\alpha_{0})$ with $\\alpha_{0}=x_{0}+y_{0}\\omega\\in\\mathcal O_{K}$.  \n\nLet $p$ be inert, i.e.\\ $\\bigl(\\tfrac{17}{p}\\bigr)=-1$.  \nThen $(p)$ remains prime in $\\mathcal O_{K}$, say $(p)=\\mathfrak P=\\bar{\\mathfrak P}$, and for every $\\beta\\in\\mathcal O_{K}$ we have  \n\\[\nv_{p}\\bigl(N_{K/\\mathbb Q}(\\beta)\\bigr)\n      =2\\,v_{\\mathfrak P}(\\beta)\n\\]\nso is even.  Applying this to $\\beta=\\alpha_{0}$ gives  \n$v_{p}(N)=e_{p}$ even, establishing (P).\n\nStep 2.  Sufficiency of (P).  \n\n2.1  Ideal factorisation.  \nWrite  \n\\[\nN=\\Bigl(\\prod_{i\\in I}p_{i}^{e_{i}}\\Bigr)\n   \\Bigl(\\prod_{j\\in I^{c}}p_{j}^{e_{j}}\\Bigr).\n\\]\nFor every $j\\in I^{c}$ choose one of the two prime ideals above $p_{j}$ and\ncall it $\\mathfrak p_{j}$.  \nFor $i\\in I$ the only prime ideal is $\\mathfrak p_{i}=(p_{i})$.  \nBecause $e_{i}$ is even, put $e_{i}=2m_{i}$.  Set  \n\\[\n\\mathfrak a:=\\prod_{i\\in I}\\mathfrak p_{i}^{\\,m_{i}}\n            \\prod_{j\\in I^{c}}\\mathfrak p_{j}^{\\,e_{j}},\n\\qquad\n(N)=\\mathfrak a\\,\\bar{\\mathfrak a}.\n\\tag{1}\n\\]\n\n2.2  Principality in the \\emph{narrow} class group.  \nSince $h^{+}(K)=1$, the ideal $\\mathfrak a$ is generated by a\n\\emph{totally positive} element; choose $\\beta\\in\\mathcal O_{K}$ with  \n\\[\n\\mathfrak a=(\\beta),\\qquad\n\\beta\\succ 0.\n\\]\nTaking norms in (1) gives  \n\\[\nN_{K/\\mathbb Q}(\\beta)=u\\cdot |N|,\\qquad u\\in\\mathcal O_{K}^{\\times},\\quad\nN_{K/\\mathbb Q}(u)=\\pm1 .\n\\]\nBecause $\\beta$ is totally positive, its norm is positive, forcing\n$N_{K/\\mathbb Q}(u)=1$.  Hence  \n\\[\nN_{K/\\mathbb Q}(\\beta)=|N|.\n\\tag{2}\n\\]\n\n2.3  Adjusting the sign.  \nLet $\\tau:=4+\\sqrt{17}$; then $N_{K/\\mathbb Q}(\\tau)=-1$.  Define  \n\\[\n\\alpha=\n\\begin{cases}\n\\beta &\\text{if }N>0,\\\\[6pt]\n\\tau\\beta &\\text{if }N<0.\n\\end{cases}\n\\]\nIn both cases $\\alpha\\in\\mathcal O_{K}$ and  \n\\[\nN_{K/\\mathbb Q}(\\alpha)=N.\n\\tag{3}\n\\]\n\n2.4  Returning to quadratic forms.  \nWrite $\\alpha=x+y\\omega$ with $x,y\\in\\mathbb Z$.  \nBy (0) and (3) we have $g(x,y)=N$, whence  \n\\[\nf(x-y,y)=N,\n\\]\nso $(x-y,y)$ is an integral solution.  \nThus (P) $\\Longrightarrow$ solvability, completing the proof of $(\\diamondsuit)$.\n\nStep 3.  Local solvability and the Hasse–Minkowski principle.  \n\n3.1  $(\\diamondsuit)\\Rightarrow$ local solvability.  \nAn integral solution gives one modulo every $p^{k}$.\n\n3.2  Local solvability $\\Longrightarrow$ (P).  \nAssume that \\emph{for every prime power $p^{k}$} the congruence  \n\\[\nf(x,y)\\equiv N\\pmod{p^{k}}\n\\tag{4}\n\\]\nis solvable.  Fix an inert prime $p$ and pass to the\n$p$–adic completion.  As $f$ is unimodularly equivalent to the norm form\n$g$, (4) yields an element $\\alpha\\in\\mathcal O_{K}\\otimes\\mathbb Z_{p}$\nwith  \n\\[\nN_{K/\\mathbb Q}(\\alpha)=N.\n\\tag{5}\n\\]\n\nLet $v_{p}$ be the $p$–adic valuation on $\\mathbb Q_{p}$ and extend it\nto $K\\otimes\\mathbb Q_{p}$ (there is a \\emph{unique} extension because\n$p$ is unramified in $K$).  For every $\\beta$ in this local field\n\\[\nv_{p}\\bigl(N_{K/\\mathbb Q}(\\beta)\\bigr)=2\\,v_{p}(\\beta).\n\\tag{6}\n\\]\nIndeed, the residue field extension is still degree $2$, hence unramified,\nso the uniformiser of $\\mathcal O_{K}\\otimes\\mathbb Z_{p}$ is a uniformiser\nof $\\mathbb Z_{p}$, and the norm multiplies valuations by $2$.\n\nApplying (6) to $\\alpha$ in (5) we get  \n\\[\nv_{p}(N)=2\\,v_{p}(\\alpha)\\equiv 0\\pmod 2 .\n\\]\nTherefore $e_{p}=v_{p}(N)$ is even for every inert $p$, i.e.\\ (P) holds.\n\nConsequently “complete local solvability'' is equivalent to (P).\nTogether with Step 2 this proves that $f$ satisfies the\nHasse–Minkowski principle.\n\nStep 4.  Application to $N=122=2\\cdot 61$.  \nCompute  \n\\[\n\\left(\\dfrac{17}{2}\\right)=1,\\qquad\n\\left(\\dfrac{17}{61}\\right)=\n\\left(\\dfrac{61}{17}\\right)=\n\\left(\\dfrac{10}{17}\\right)=\n\\left(\\dfrac{2}{17}\\right)\\left(\\dfrac{5}{17}\\right)=1\\cdot(-1)=-1.\n\\]\nHence $61$ is inert and occurs with exponent $1$, violating (P).  \nBy $(\\diamondsuit)$ the equation  \n\\[\nx^{2}+3xy-2y^{2}=122\n\\]\nhas \\emph{no} integer solutions.\n\nStep 5.  The “one–bit’’ obstruction.  \nFrom Step 1, if $f$ represents $N$ then every inert prime occurs with an\n\\emph{even} exponent, so  \n\\[\n\\left(\\dfrac{N}{17}\\right)\n   =\\prod_{i\\in I}\\left(\\dfrac{17}{p_{i}}\\right)^{e_{i}}\n   =(-1)^{\\sum_{i\\in I}e_{i}}=1 .\n\\]\nThus an odd \\emph{total} number of inert prime factors already prevents\nsolvability, reproducing the classical criterion.  However this is\nstrictly weaker than (P): parity has to be checked \\emph{prime by prime}\n(e.g.\\ $N=15=3\\cdot5$ fails $(\\diamondsuit)$ although the total number\nof inert primes is even).\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.394298",
        "was_fixed": false,
        "difficulty_analysis": "•  The original (and kernel) problems required only a single congruence\n   argument modulo one prime; the enhanced variant demands a full local–global\n   characterisation of the values taken by the quadratic form.\n\n•  To prove (*), one must\n   – transform a binary quadratic form into a norm form,  \n   – work in the quadratic integer ring ℤ[√17],  \n   – use the unit group of that ring (via the Pell equation) to lift local\n     information to an integral solution, i.e. employ Dirichlet’s “Pell\n     trick’’/Hasse–Minkowski type reasoning,  \n   – combine these observations with quadratic reciprocity.\n\n•  Part (**) intertwines congruence conditions modulo 4 and 17, demanding\n   simultaneous control of two quadratic symbols and multiplicativity; this\n   yields an infinite family of forbidden integers rather than a single\n   numerical instance.\n\n•  Altogether the enhanced problem obliges the solver to master\n   local solvability, norm forms, the structure of units in real quadratic\n   fields, the multiplicativity of Legendre symbols, and quadratic reciprocity\n   – a substantial escalation in both technical depth and conceptual\n   breadth compared with the single-step residue-check of the kernel task."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}