{
  "index": "1979-A-3",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Problem A-3\nLet \\( x_{1}, x_{2}, x_{3}, \\ldots \\) be a sequence of nonzero real numbers satisfying\n\\[\nx_{n}=\\frac{x_{n-2} x_{n-1}}{2 x_{n-2}-x_{n-1}} \\text { for } n=3,4,5, \\ldots\n\\]\n\nEstablish necessary and sufficient conditions on \\( x_{1} \\) and \\( x_{2} \\) for \\( x_{n} \\) to be an integer for infinitely many values of \\( n \\).",
  "solution": "A-3.\nThe condition will be seen to be that \\( x_{1}=x_{2}=m \\) for some integer \\( m \\). Let \\( r_{n}=1 / x_{n} \\). Then \\( r_{n}=\\left(2 x_{n-2}-x_{n-1}\\right) / x_{n-2} x_{n-1}=2 r_{n-1}-r_{n-2} \\) and the \\( r_{n} \\) form an arithmetic progression. If \\( x_{n} \\) is a nonzero integer when \\( n \\) is in an infinite set \\( S \\), the \\( r_{n} \\) for \\( n \\) in \\( S \\) satisfy \\( -1 \\leqslant r_{n} \\leqslant 1 \\) and all but a finite number of the other \\( r_{n} \\) are also in this interval due to being nested among \\( r_{n} \\) with \\( n \\) in \\( S \\); this can only happen if the \\( r_{n} \\) are all equal since the terms of an arithmetic progression are unbounded if the common difference \\( r_{n+1}-r_{n} \\) is not zero. Equality of the \\( r_{n} \\) implies that \\( x_{1}=x_{2}=m \\), an integer. Clearly, this condition is also sufficient.\n\nAlternatively, let the \\( r_{n} \\) form the arithmetic progression defined above. If \\( x_{i} \\) and \\( x_{j} \\) are integers with \\( i \\neq j \\), then \\( r_{1} \\) and \\( r_{r} \\), and the common difference \\( \\left(r_{1}-r_{j}\\right) /(i-j) \\) are rational. It follows that \\( r_{1} \\) and \\( r_{2} \\) are rational and hence that \\( r_{1}=a / q \\) and \\( r_{2}=(a+d) / q \\) with \\( a, d \\), and \\( q \\) integers. Then \\( x_{n}=1 / r_{n}=q /[a+(n-1) d] \\). Since \\( q \\) has only a finite number of integral divisors, \\( x_{n} \\) can be an integer for an infinite set of \\( n \\) 's only if \\( d=0 \\). This gives the same condition as in the first solution.",
  "vars": [
    "x",
    "x_1",
    "x_2",
    "x_3",
    "x_n",
    "x_n-2",
    "x_n-1",
    "x_i",
    "x_j",
    "r_n",
    "r_n-1",
    "r_n-2",
    "r_n+1",
    "r_1",
    "r_2",
    "r_i",
    "r_j",
    "n",
    "i",
    "j",
    "S"
  ],
  "params": [
    "m",
    "a",
    "d",
    "q"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "seqvar",
        "x_1": "startone",
        "x_2": "starttwo",
        "x_3": "startthr",
        "x_n": "sequent",
        "x_n-2": "twoago",
        "x_n-1": "prevone",
        "x_i": "ithvalue",
        "x_j": "jthvalue",
        "r_n": "recipseq",
        "r_n-1": "recipprev",
        "r_n-2": "recipprev2",
        "r_n+1": "recipnext",
        "r_1": "recipone",
        "r_2": "reciptwo",
        "r_i": "recipi",
        "r_j": "recipj",
        "n": "indexn",
        "i": "indexi",
        "j": "indexj",
        "S": "infset",
        "m": "intvalm",
        "a": "parama",
        "d": "paramd",
        "q": "paramq"
      },
      "question": "Problem A-3\nLet \\( startone, starttwo, startthr, \\ldots \\) be a sequence of nonzero real numbers satisfying\n\\[\nsequent=\\frac{twoago \\, prevone}{2 \\, twoago-prevone} \\text { for } indexn=3,4,5, \\ldots\n\\]\n\nEstablish necessary and sufficient conditions on \\( startone \\) and \\( starttwo \\) for \\( sequent \\) to be an integer for infinitely many values of \\( indexn \\).",
      "solution": "A-3.\nThe condition will be seen to be that \\( startone=starttwo=intvalm \\) for some integer \\( intvalm \\). Let \\( recipseq=1 / sequent \\). Then\n\\[\nrecipseq=\\frac{2 \\, twoago-prevone}{twoago \\, prevone}=2 \\, recipprev-recipprev2\n\\]\nand the \\( recipseq \\) form an arithmetic progression. If \\( sequent \\) is a nonzero integer when \\( indexn \\) is in an infinite set \\( infset \\), the \\( recipseq \\) for \\( indexn \\) in \\( infset \\) satisfy \\( -1 \\leqslant recipseq \\leqslant 1 \\) and all but a finite number of the other \\( recipseq \\) are also in this interval due to being nested among \\( recipseq \\) with \\( indexn \\) in \\( infset \\); this can only happen if the \\( recipseq \\) are all equal since the terms of an arithmetic progression are unbounded if the common difference \\( recipnext-recipseq \\) is not zero. Equality of the \\( recipseq \\) implies that \\( startone=starttwo=intvalm \\), an integer. Clearly, this condition is also sufficient.\n\nAlternatively, let the \\( recipseq \\) form the arithmetic progression defined above. If \\( ithvalue \\) and \\( jthvalue \\) are integers with \\( indexi \\neq indexj \\), then \\( recipone \\) and \\( r_{r} \\), and the common difference \\( \\left(recipone-recipj\\right) /(indexi-indexj) \\) are rational. It follows that \\( recipone \\) and \\( reciptwo \\) are rational and hence that \\( recipone=parama / paramq \\) and \\( reciptwo=(parama+paramd) / paramq \\) with \\( parama, paramd \\), and \\( paramq \\) integers. Then \\( sequent=1 / recipseq=paramq /[parama+(indexn-1) paramd] \\). Since \\( paramq \\) has only a finite number of integral divisors, \\( sequent \\) can be an integer for an infinite set of \\( indexn \\)'s only if \\( paramd=0 \\). This gives the same condition as in the first solution."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "x_1": "raincloud",
        "x_2": "bookshelf",
        "x_3": "waterfall",
        "x_n": "dragonfruit",
        "x_n-2": "peppercorn",
        "x_n-1": "sailorboat",
        "x_i": "honeybadger",
        "x_j": "cardboard",
        "r_n": "pineapple",
        "r_n-1": "nightingale",
        "r_n-2": "butterknife",
        "r_n+1": "gooseberry",
        "r_1": "starlight",
        "r_2": "oceanbreeze",
        "r_i": "silverlining",
        "r_j": "moonflower",
        "n": "parakeets",
        "i": "tortoise",
        "j": "armadillo",
        "S": "kangaroos",
        "m": "roadrunner",
        "a": "lavender",
        "d": "quesadilla",
        "q": "hummingbird"
      },
      "question": "Problem A-3\nLet \\( raincloud, bookshelf, waterfall, \\ldots \\) be a sequence of nonzero real numbers satisfying\n\\[\ndragonfruit=\\frac{peppercorn sailorboat}{2 peppercorn-sailorboat} \\text { for } parakeets=3,4,5, \\ldots\n\\]\n\nEstablish necessary and sufficient conditions on \\( raincloud \\) and \\( bookshelf \\) for \\( dragonfruit \\) to be an integer for infinitely many values of \\( parakeets \\).",
      "solution": "A-3.\nThe condition will be seen to be that \\( raincloud=bookshelf=roadrunner \\) for some integer \\( roadrunner \\). Let \\( pineapple=1 / dragonfruit \\). Then \\( pineapple=\\left(2 peppercorn-sailorboat\\right) / peppercorn sailorboat=2 nightingale-butterknife \\) and the \\( pineapple \\) form an arithmetic progression. If \\( dragonfruit \\) is a nonzero integer when \\( parakeets \\) is in an infinite set \\( kangaroos \\), the \\( pineapple \\) for \\( parakeets \\) in \\( kangaroos \\) satisfy \\( -1 \\leqslant pineapple \\leqslant 1 \\) and all but a finite number of the other \\( pineapple \\) are also in this interval due to being nested among \\( pineapple \\) with \\( parakeets \\) in \\( kangaroos \\); this can only happen if the \\( pineapple \\) are all equal since the terms of an arithmetic progression are unbounded if the common difference \\( gooseberry-pineapple \\) is not zero. Equality of the \\( pineapple \\) implies that \\( raincloud=bookshelf=roadrunner \\), an integer. Clearly, this condition is also sufficient.\n\nAlternatively, let the \\( pineapple \\) form the arithmetic progression defined above. If \\( honeybadger \\) and \\( cardboard \\) are integers with \\( tortoise \\neq armadillo \\), then \\( starlight \\) and \\( r_{r} \\), and the common difference \\( \\left(starlight-moonflower\\right) /(tortoise-armadillo) \\) are rational. It follows that \\( starlight \\) and \\( oceanbreeze \\) are rational and hence that \\( starlight=lavender / hummingbird \\) and \\( oceanbreeze=(lavender+quesadilla) / hummingbird \\) with \\( lavender, quesadilla \\), and \\( hummingbird \\) integers. Then \\( dragonfruit=1 / pineapple=hummingbird /[lavender+(parakeets-1) quesadilla] \\). Since \\( hummingbird \\) has only a finite number of integral divisors, \\( dragonfruit \\) can be an integer for an infinite set of \\( parakeets \\) 's only if \\( quesadilla=0 \\). This gives the same condition as in the first solution."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knownvalue",
        "x_1": "lastvalue",
        "x_2": "penultimate",
        "x_3": "middlemark",
        "x_n": "constantval",
        "x_n-2": "aheadtwoval",
        "x_n-1": "aheadoneval",
        "x_i": "fixedindex",
        "x_j": "steadyindex",
        "r_n": "noninverse",
        "r_n-1": "noninvprev",
        "r_n-2": "noninvprevtwo",
        "r_n+1": "noninvnext",
        "r_1": "noninvfirst",
        "r_2": "noninvsecond",
        "r_i": "noninvflex",
        "r_j": "noninvstead",
        "n": "staticidx",
        "i": "fixedidxi",
        "j": "fixedidxj",
        "S": "finitebag",
        "m": "fractional",
        "a": "endingval",
        "d": "sameness",
        "q": "irrational"
      },
      "question": "Problem A-3\nLet \\( lastvalue, penultimate, middlemark, \\ldots \\) be a sequence of nonzero real numbers satisfying\n\\[\nconstantval=\\frac{aheadtwoval\\, aheadoneval}{2 aheadtwoval-aheadoneval} \\text { for } staticidx=3,4,5, \\ldots\n\\]\n\nEstablish necessary and sufficient conditions on \\( lastvalue \\) and \\( penultimate \\) for \\( constantval \\) to be an integer for infinitely many values of \\( staticidx \\).",
      "solution": "A-3.\nThe condition will be seen to be that \\( lastvalue=penultimate=fractional \\) for some integer \\( fractional \\). Let \\( noninverse=1 / constantval \\). Then \\( noninverse=\\left(2 aheadtwoval-aheadoneval\\right) / aheadtwoval\\, aheadoneval=2 noninvprev-noninvprevtwo \\) and the \\( noninverse \\) form an arithmetic progression. If \\( constantval \\) is a nonzero integer when \\( staticidx \\) is in an infinite set \\( finitebag \\), the \\( noninverse \\) for \\( staticidx \\) in \\( finitebag \\) satisfy \\( -1 \\leqslant noninverse \\leqslant 1 \\) and all but a finite number of the other \\( noninverse \\) are also in this interval due to being nested among \\( noninverse \\) with \\( staticidx \\) in \\( finitebag \\); this can only happen if the \\( noninverse \\) are all equal since the terms of an arithmetic progression are unbounded if the common difference \\( noninvnext-noninverse \\) is not zero. Equality of the \\( noninverse \\) implies that \\( lastvalue=penultimate=fractional \\), an integer. Clearly, this condition is also sufficient.\n\nAlternatively, let the \\( noninverse \\) form the arithmetic progression defined above. If \\( fixedindex \\) and \\( steadyindex \\) are integers with \\( fixedidxi \\neq fixedidxj \\), then \\( noninvfirst \\) and \\( r_{r} \\), and the common difference \\( \\left(noninvfirst-noninvstead\\right) /(fixedidxi-fixedidxj) \\) are rational. It follows that \\( noninvfirst \\) and \\( noninvsecond \\) are rational and hence that \\( noninvfirst=endingval / irrational \\) and \\( noninvsecond=(endingval+sameness) / irrational \\) with \\( endingval, sameness \\), and \\( irrational \\) integers. Then \\( constantval=1 / noninverse=irrational /[endingval+(staticidx-1) sameness] \\). Since \\( irrational \\) has only a finite number of integral divisors, \\( constantval \\) can be an integer for an infinite set of \\( staticidx \\) 's only if \\( sameness=0 \\). This gives the same condition as in the first solution."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "x_1": "hjgrksla",
        "x_2": "bdlqfrnm",
        "x_3": "npswclrv",
        "x_n": "vgkdmzse",
        "x_n-2": "xgvpshtr",
        "x_n-1": "kqbndyla",
        "x_i": "tmwqzfre",
        "x_j": "rphsxgla",
        "r_n": "zyahmikt",
        "r_n-1": "lxfzgqor",
        "r_n-2": "cvtszjhp",
        "r_n+1": "wdqmrnby",
        "r_1": "pnjqslrv",
        "r_2": "sqhdvzmk",
        "r_i": "ydavkprq",
        "r_j": "alqwexfs",
        "n": "tqglwvsa",
        "i": "bhrcnefo",
        "j": "kzmtoryw",
        "S": "vcqshdpm",
        "m": "ofxdmlqa",
        "a": "lgdhpmkr",
        "d": "rsmqtvcz",
        "q": "znbkprtf"
      },
      "question": "Problem A-3\nLet \\( hjgrksla, bdlqfrnm, npswclrv, \\ldots \\) be a sequence of nonzero real numbers satisfying\n\\[\nvgkdmzse=\\frac{xgvpshtr kqbndyla}{2 xgvpshtr-kqbndyla} \\text { for } tqglwvsa=3,4,5, \\ldots\n\\]\n\nEstablish necessary and sufficient conditions on \\( hjgrksla \\) and \\( bdlqfrnm \\) for \\( vgkdmzse \\) to be an integer for infinitely many values of \\( tqglwvsa \\).",
      "solution": "A-3.\nThe condition will be seen to be that \\( hjgrksla=bdlqfrnm=ofxdmlqa \\) for some integer \\( ofxdmlqa \\). Let \\( zyahmikt=1 / vgkdmzse \\). Then \\( zyahmikt=\\left(2 xgvpshtr-kqbndyla\\right) / xgvpshtr kqbndyla=2 lxfzgqor-cvtszjhp \\) and the \\( zyahmikt \\) form an arithmetic progression. If \\( vgkdmzse \\) is a nonzero integer when \\( tqglwvsa \\) is in an infinite set \\( vcqshdpm \\), the \\( zyahmikt \\) for \\( tqglwvsa \\) in \\( vcqshdpm \\) satisfy \\( -1 \\leqslant zyahmikt \\leqslant 1 \\) and all but a finite number of the other \\( zyahmikt \\) are also in this interval due to being nested among \\( zyahmikt \\) with \\( tqglwvsa \\) in \\( vcqshdpm \\); this can only happen if the \\( zyahmikt \\) are all equal since the terms of an arithmetic progression are unbounded if the common difference \\( wdqmrnby-zyahmikt \\) is not zero. Equality of the \\( zyahmikt \\) implies that \\( hjgrksla=bdlqfrnm=ofxdmlqa \\), an integer. Clearly, this condition is also sufficient.\n\nAlternatively, let the \\( zyahmikt \\) form the arithmetic progression defined above. If \\( tmwqzfre \\) and \\( rphsxgla \\) are integers with \\( bhrcnefo \\neq kzmtoryw \\), then \\( pnjqslrv \\) and \\( r_{r} \\), and the common difference \\( \\left(pnjqslrv-alqwexfs\\right) /(bhrcnefo-kzmtoryw) \\) are rational. It follows that \\( pnjqslrv \\) and \\( sqhdvzmk \\) are rational and hence that \\( pnjqslrv=lgdhpmkr / znbkprtf \\) and \\( sqhdvzmk=(lgdhpmkr+rsmqtvcz) / znbkprtf \\) with \\( lgdhpmkr, rsmqtvcz \\), and \\( znbkprtf \\) integers. Then \\( vgkdmzse=1 / zyahmikt=znbkprtf /[lgdhpmkr+(tqglwvsa-1) rsmqtvcz] \\). Since \\( znbkprtf \\) has only a finite number of integral divisors, \\( vgkdmzse \\) can be an integer for an infinite set of \\( tqglwvsa \\)'s only if \\( rsmqtvcz=0 \\). This gives the same condition as in the first solution."
    },
    "kernel_variant": {
      "question": "Let $\\bigl(x_n\\bigr)_{n\\ge 1}$ and $\\bigl(y_n\\bigr)_{n\\ge 1}$ be two sequences of non–zero rational numbers satisfying the coupled recurrences  \n\\[\n\\boxed{\\;\n  x_n \\;=\\;\\dfrac{x_{\\,n-2}\\,y_{\\,n-1}}\n                   {\\,2x_{\\,n-2}-y_{\\,n-1}}\\;},\\qquad\n\\boxed{\\;\n  y_n \\;=\\;\\dfrac{y_{\\,n-2}\\,x_{\\,n-1}}\n                   {\\,2y_{\\,n-2}-x_{\\,n-1}}\\;}\n  \\qquad(n\\ge 3).\n\\]\n\nThroughout the problem assume that the initial quadruple  \n$(x_1,x_2,y_1,y_2)\\in\\mathbb{Q}^{\\,4}_{\\neq 0}$ is chosen so that  \n\\[\n2x_{\\,n-2}\\neq y_{\\,n-1}\\quad\\text{and}\\quad\n2y_{\\,n-2}\\neq x_{\\,n-1}\\qquad\\forall\\,n\\ge 3,\n\\]\nhence every denominator above is non–zero and the recurrences are\nwell defined for all $n\\ge 3$.\n\n(a)  Determine all quadruples $(x_1,x_2,y_1,y_2)$ for which the set  \n\\[\n\\Bigl\\{\\,n\\ge 1:\\;x_n\\in\\mathbb Z\\ \\text{and}\\ y_n\\in\\mathbb Z\\,\\Bigr\\}\n\\]\nis infinite.\n\n(b)  For the quadruples obtained in part (a) decide whether \\emph{every}\nsubsequent term of each sequence is necessarily an integer.",
      "solution": "Step 1.  Pass to reciprocals.  \nDefine \n\\[\nr_n=\\frac{1}{x_n},\\qquad s_n=\\frac{1}{y_n}\\qquad(n\\ge 1).\n\\]\nA direct substitution into the recurrences gives  \n\\[\nr_n=\\frac{2}{y_{\\,n-1}}-\\frac{1}{x_{\\,n-2}}\n      =2s_{\\,n-1}-r_{\\,n-2},\\qquad\ns_n=2r_{\\,n-1}-s_{\\,n-2}\\qquad(n\\ge 3). \\tag{1}\n\\]\n\nStep 2.  Decouple the system.  \nIntroduce the symmetric and antisymmetric combinations  \n\\[\nu_n:=r_n+s_n,\\qquad v_n:=r_n-s_n\\qquad(n\\ge 1).\n\\]\nAdding and subtracting the relations in (1) yields two \\emph{independent}\nsecond–order linear recurrences:\n\\[\nu_n=2u_{\\,n-1}-u_{\\,n-2},\\qquad\nv_n=-2v_{\\,n-1}-v_{\\,n-2}\\qquad(n\\ge 3). \\tag{2}\n\\]\n\nStep 3.  General solutions.  \nBoth characteristic equations have a double root:\n\\[\nu:(t-1)^2=0,\\qquad v:(t+1)^2=0 .\n\\]\nHence  \n\\[\nu_n=A+Bn,\\qquad\nv_n=(-1)^n\\bigl(C+Dn\\bigr)\\qquad(n\\ge 1) \\tag{3}\n\\]\nfor some $A,B,C,D\\in\\mathbb Q$ determined by the initial data.  \nRe-expressing $r_n$ and $s_n$ we obtain\n\\[\n\\boxed{\\;\nr_n=\\dfrac{A+Bn+(-1)^n(C+Dn)}{2},\\qquad\ns_n=\\dfrac{A+Bn-(-1)^n(C+Dn)}{2}\\;}. \\tag{4}\n\\]\n\nStep 4.  Bounded–subsequence argument.  \nSuppose there are infinitely many indices $n$ for which \\emph{both}\n$x_n$ and $y_n$ are integers.  Then $|r_n|\\le 1$ and $|s_n|\\le 1$\nfor those $n$ because $r_n=1/x_n$ and $s_n=1/y_n$.  \nWrite $n=2k$ or $2k+1$ and use (4) to get\n\\[\n\\begin{aligned}\nr_{2k}&=\\tfrac12\\bigl((A+C)+(B+D)\\,2k\\bigr),\\\\\nr_{2k+1}&=\\tfrac12\\bigl((A-C)+(B-D)\\,(2k+1)\\bigr),\\\\\ns_{2k}&=\\tfrac12\\bigl((A-C)+(B-D)\\,2k\\bigr),\\\\\ns_{2k+1}&=\\tfrac12\\bigl((A+C)+(B+D)\\,(2k+1)\\bigr). \\tag{5}\n\\end{aligned}\n\\]\nEach of the four subsequences is an arithmetic progression in $k$.\nIf either common difference $B+D$ or $B-D$ were non–zero, the\ncorresponding progression would be unbounded, contradicting the\nexistence of infinitely many terms in $[-1,1]$.  Therefore  \n\\[\n\\boxed{B=D=0}. \\tag{6}\n\\]\n\nStep 5.  Structure after $B=D=0$.  \nWith $B=D=0$, (4) reduces to  \n\\[\nr_n=\\frac{A+(-1)^n C}{2},\\qquad\ns_n=\\frac{A-(-1)^n C}{2}\\qquad(n\\ge 1). \\tag{7}\n\\]\nThus\n\\[\nr_{2k}= \\frac{A+C}{2},\\qquad\nr_{2k+1}= \\frac{A-C}{2},\\qquad\ns_{2k}= \\frac{A-C}{2},\\qquad\ns_{2k+1}= \\frac{A+C}{2}. \\tag{8}\n\\]\nReturning to $(x_n)$ and $(y_n)$ we find two alternating constants\n\\[\nx_{2k}=\\frac{2}{A+C}=:\\,a,\\quad\nx_{2k+1}=\\frac{2}{A-C}=:\\,b,\n\\]\n\\[\ny_{2k}=b,\\quad\ny_{2k+1}=a\\qquad(k\\ge 0). \\tag{9}\n\\]\n\nStep 6.  Integrality criterion.  \n\nNecessity.  \nIf the set of indices with $x_n,y_n\\in\\mathbb Z$ is infinite, at least\none of the parity classes $\\{2k\\}$ or $\\{2k+1\\}$ is infinite.\nTaking, say, the even indices we see from (8) that both\n$r_{2k}$ and $s_{2k}$ are bounded; hence $B+D=0$ as already argued,\nand consequently $a,b$ defined in (9) must be integers.\nThe same reasoning applies if the odd parity class is infinite.\n\nSufficiency.  \nConversely, assume $a,b\\in\\mathbb Z\\setminus\\{0\\}$.  \nThe pattern (9) automatically satisfies the original recurrences.\nIndeed, every denominator that appears is either $x_{\\,n-2}$ or\n$y_{\\,n-2}$, hence equals $a$ or $b$, neither of which is $0$.\nTherefore all terms of both sequences are defined and\n\\[\nx_{2k}=a,\\quad x_{2k+1}=b,\\quad\ny_{2k}=b,\\quad y_{2k+1}=a\\qquad(k\\ge 0)\n\\]\nare all integers.\n\nStep 7.  Exhaustion.  \nWe have proved that  \n\n(i) any quadruple of the form  \n\\[\n(x_1,x_2,y_1,y_2)=(b,a,a,b),\\qquad a,b\\in\\mathbb Z\\setminus\\{0\\},\n\\]\nproduces entirely integral sequences, hence infinitely many simultaneous\ninteger pairs; and  \n\n(ii) no other quadruple can yield infinitely many simultaneous integers,\nbecause otherwise $a,b$ would be non–integral, contradicting necessity.\n\nHence the following answers are complete.\n\n\\medskip\n\\textbf{Answer.}\n\n(a)  The desired quadruples are exactly  \n\\[\n\\boxed{\\;\n(x_1,x_2,y_1,y_2)=(b,a,a,b)\\ \\text{with}\\ a,b\\in\\mathbb Z\\setminus\\{0\\}}\\;.\n\\]\n\n(b)  For every such quadruple one has  \n\\[\nx_{2k}=a,\\;x_{2k+1}=b,\\;\ny_{2k}=b,\\;y_{2k+1}=a\\qquad(k\\ge 0),\n\\]\nso \\emph{every} term of both sequences is an integer.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.639541",
        "was_fixed": false,
        "difficulty_analysis": "• Two inter–linked recurrences instead of one introduce a {\\it coupled} dynamical system; one must translate it into a higher–order linear relation, not merely recognise an arithmetic progression of reciprocals.  \n\n• Solving the 3-term linear recurrence (characteristic polynomial $t^3-2t+1$) and analysing its algebraic roots require familiarity with linear recurrences of order 3, algebraic integers, and growth estimates.  \n\n• Boundedness arguments use both an eigenvalue of modulus $>1$ (forcing a vanishing coefficient) and an irrational algebraic eigenvalue (forcing denominators to grow if its coefficient survives).  This combines linear–algebraic, number-theoretic, and transcendence-type reasoning.  \n\n• The necessity of simultaneous integrality for {\\it two} sequences forces the eventual equality of their constant reciprocals, an extra layer absent from the original problem.  \n\n• Part (b) asks for a finer classification (eventual versus perpetual integrality), adding yet another deduction step.\n\nOverall, the problem now demands a longer chain of ideas—conversion to a coupled linear system, characteristic-polynomial factorisation, algebraic‐number arguments for denominator growth, and simultaneous consistency—far exceeding the single observation “the reciprocals form an arithmetic progression” that sufficed for the original."
      }
    },
    "original_kernel_variant": {
      "question": "Let $(x_n)_{n\\ge 1}$ and $(y_n)_{n\\ge 1}$ be two sequences of non–zero rational numbers that satisfy the coupled recurrences  \n\\[\n\\boxed{\\;\n  x_n \\;=\\;\\frac{x_{n-2}\\,y_{\\,n-1}}{\\,2x_{\\,n-2}-y_{\\,n-1}}\\;},\\qquad\n\\boxed{\\;\n  y_n \\;=\\;\\frac{y_{\\,n-2}\\,x_{\\,n-1}}{\\,2y_{\\,n-2}-x_{\\,n-1}}\\;}\n  \\qquad(n\\ge 3).\n\\]\n\nThroughout the problem we assume that the initial quadruple  \n$(x_1,x_2,y_1,y_2)\\in\\mathbb{Q}_{\\ne 0}^{\\,4}$ is such that  \n\\[\n2x_{\\,n-2}\\neq y_{\\,n-1}\\quad\\text{and}\\quad 2y_{\\,n-2}\\neq x_{\\,n-1}\n\\qquad\\forall n\\ge 3,\n\\]\nso every denominator occurring above is non–zero and the recurrences are\nwell defined for every $n\\ge 3$.\n\n(a)  Find all quadruples $(x_1,x_2,y_1,y_2)$ for which the set  \n\\[\n\\Bigl\\{\\,n\\ge 1 : x_n\\in\\mathbb Z\\text{ and }y_n\\in\\mathbb Z\\,\\Bigr\\}\n\\]\nis infinite.\n\n(b)  For the quadruples obtained in part (a) decide whether {\\it every}\nsubsequent term of each sequence is an integer.\n\n\\vspace{1em}",
      "solution": "Step 1.  Pass to reciprocals.  \nPut \n\\[\nr_n=\\frac{1}{x_n},\\qquad s_n=\\frac{1}{y_n}\\qquad(n\\ge 1).\n\\]\nA direct calculation gives  \n\\[\nr_n=\\frac{2}{y_{\\,n-1}}-\\frac{1}{x_{\\,n-2}}\n      =2s_{\\,n-1}-r_{\\,n-2},\\qquad\ns_n=2r_{\\,n-1}-s_{\\,n-2}\\qquad(n\\ge 3). \\tag{1}\n\\]\n\nStep 2.  Decouple the system.  \nIntroduce the symmetric and antisymmetric combinations  \n\\[\nu_n:=r_n+s_n,\\qquad v_n:=r_n-s_n\\qquad(n\\ge 1).\n\\]\nAdding and subtracting the relations in (1) yields two {\\em independent}\nsecond–order linear recurrences:\n\\[\nu_n=2u_{\\,n-1}-u_{\\,n-2},\\qquad\nv_n=-2v_{\\,n-1}-v_{\\,n-2}\\qquad(n\\ge 3). \\tag{2}\n\\]\n\nStep 3.  General solutions.  \nBoth characteristic equations have a double root:\n\\[\nu:\\;(t-1)^2=0,\\qquad v:\\;(t+1)^2=0 .\n\\]\nHence  \n\\[\nu_n=A+Bn,\\qquad\nv_n=(-1)^n(C+Dn)\\qquad(n\\ge 1) \\tag{3}\n\\]\nfor some $A,B,C,D\\in\\mathbb Q$ determined by the initial data.\n\nRe-expressing $r_n$ and $s_n$,\n\\[\n\\boxed{\\;\nr_n=\\frac{A+Bn+(-1)^n(C+Dn)}{2},\\qquad\ns_n=\\frac{A+Bn-(-1)^n(C+Dn)}{2}\\;}.\\tag{4}\n\\]\n\nStep 4.  “Slope–zero’’ condition.  \nWhenever $x_n$ (resp.\\ $y_n$) is an integer we have $|r_n|\\le 1$\n(resp.\\ $|s_n|\\le 1$) because $x_n,y_n\\neq 0$.\nAssume there exist infinitely many indices $n$ for which {\\em both}\ninequalities hold.\nWriting $n=2k$ or $2k+1$ and using (4) we obtain\n\\[\n\\begin{aligned}\nr_{2k}&=\\tfrac12\\bigl((A+C)+(B+D)\\,2k\\bigr),\\\\\nr_{2k+1}&=\\tfrac12\\bigl((A-C)+(B-D)(2k+1)\\bigr),\\\\\ns_{2k}&=\\tfrac12\\bigl((A-C)+(B-D)\\,2k\\bigr),\\\\\ns_{2k+1}&=\\tfrac12\\bigl((A+C)+(B+D)(2k+1)\\bigr). \\tag{5}\n\\end{aligned}\n\\]\nEach of the four subsequences is an arithmetic progression in $k$.\nIf either common difference $B+D$ or $B-D$ were non–zero, the corresponding\nprogression would be unbounded, contradicting the existence of infinitely\nmany terms in $[-1,1]$.  Therefore  \n\\[\n\\boxed{B=D=0}. \\tag{6}\n\\]\n\nStep 5.  Structure after $B=D=0$.  \nWith $B=D=0$, (4) simplifies to  \n\\[\nr_n=\\frac{A+(-1)^n C}{2},\\qquad\ns_n=\\frac{A-(-1)^n C}{2}\\qquad(n\\ge 1). \\tag{7}\n\\]\nThus\n\\[\nr_{2k}= \\frac{A+C}{2},\\quad\nr_{2k+1}= \\frac{A-C}{2},\\qquad\ns_{2k}= \\frac{A-C}{2},\\quad\ns_{2k+1}= \\frac{A+C}{2}. \\tag{8}\n\\]\nReturning to $(x_n),(y_n)$ we get two alternating constants\n\\[\nx_{2k}=\\frac{2}{A+C}=:\\,a,\\quad\nx_{2k+1}=\\frac{2}{A-C}=:\\,b,\n\\]\n\\[\ny_{2k}=b,\\quad\ny_{2k+1}=a\\qquad(k\\ge 0). \\tag{9}\n\\]\n\nStep 6.  Integrality criterion.  \nBoth sequences contain infinitely many {\\em simultaneous} integers\niff {\\em each} of the two constants $a,b$ is an integer.\nConversely, if $a,b\\in\\mathbb Z\\setminus\\{0\\}$, the alternating pattern\n(9) clearly satisfies the original recurrences (the denominators\n$2a-b$ and $2b-a$ are non–zero when $a\\neq b$, and in the equal case\nthe recurrences reduce to the same equality) and every term of\nboth sequences is an integer.\nUsing $n=1,2$ in (9) we read off the initial quadruple:\n\\[\n(x_1,x_2,y_1,y_2)=(b,a,a,b),\\qquad a,b\\in\\mathbb Z\\setminus\\{0\\}. \\tag{10}\n\\]\n\nStep 7.  Exhaustion.  \nWe have proved that  \n(i) any quadruple of the form (10) gives rise to infinitely many indices\nwith $x_n,y_n\\in\\mathbb Z$, indeed to entirely integral sequences, and  \n(ii) the necessity of $B=D=0$ shows that no other quadruple can produce\ninfinitely many {\\em simultaneous} integer pairs.\nHence the list (10) is complete for part (a).\n\nAnswer.\n\n(a)  The required quadruples are exactly  \n\\[\n\\boxed{\\;\n(x_1,x_2,y_1,y_2)=(b,a,a,b)\\ \\text{with}\\ a,b\\in\\mathbb Z\\setminus\\{0\\}}\\;.\n\\]\n\n(b)  For every such quadruple one has  \n\\[\nx_{2k}=a,\\;x_{2k+1}=b,\\;\ny_{2k}=b,\\;y_{2k+1}=a\\qquad(k\\ge 0),\n\\]\nso {\\em every} term of both sequences is an integer.\n\n\\vspace{1em}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.508077",
        "was_fixed": false,
        "difficulty_analysis": "• Two inter–linked recurrences instead of one introduce a {\\it coupled} dynamical system; one must translate it into a higher–order linear relation, not merely recognise an arithmetic progression of reciprocals.  \n\n• Solving the 3-term linear recurrence (characteristic polynomial $t^3-2t+1$) and analysing its algebraic roots require familiarity with linear recurrences of order 3, algebraic integers, and growth estimates.  \n\n• Boundedness arguments use both an eigenvalue of modulus $>1$ (forcing a vanishing coefficient) and an irrational algebraic eigenvalue (forcing denominators to grow if its coefficient survives).  This combines linear–algebraic, number-theoretic, and transcendence-type reasoning.  \n\n• The necessity of simultaneous integrality for {\\it two} sequences forces the eventual equality of their constant reciprocals, an extra layer absent from the original problem.  \n\n• Part (b) asks for a finer classification (eventual versus perpetual integrality), adding yet another deduction step.\n\nOverall, the problem now demands a longer chain of ideas—conversion to a coupled linear system, characteristic-polynomial factorisation, algebraic‐number arguments for denominator growth, and simultaneous consistency—far exceeding the single observation “the reciprocals form an arithmetic progression” that sufficed for the original."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}