{
  "index": "1988-B-6",
  "type": "NT",
  "tag": [
    "NT",
    "ALG",
    "COMB"
  ],
  "difficulty": "",
  "question": "Prove that there exist an infinite number of ordered pairs $(a,b)$ of\nintegers such that for every positive integer $t$, the number $at+b$\nis a triangular number if and only if $t$ is a triangular number. (The\ntriangular numbers are the $t_n = n(n+1)/2$ with $n$ in $\\{0,1,2,\\dots\\}$.)\n\n\\end{itemize}\n\n\\end{document}",
  "solution": "Solution 1. It is easy to check that \\( t_{3 n+1}=9 t_{n}+1 \\), while \\( t_{3 n} \\equiv t_{3 n+2} \\equiv 0 \\) \\( (\\bmod 3) \\) for any integer \\( n \\). Hence for every positive integer \\( t, t \\) is a triangular number if and only if \\( 9 t+1 \\) is triangular. If the \\( n \\)th iterate of the linear map \\( t \\mapsto 9 t+1 \\) is \\( t \\mapsto a t+b \\), then a chain of equivalences will show that \\( t \\) is a triangular number if and only if \\( a t+b \\) is triangular. We obtain infinitely many pairs of integers \\( (a, b) \\) in this way.\n\nSolution 2. If \\( t=n(n+1) / 2 \\), then \\( 8 t+1=(2 n+1)^{2} \\). Conversely if \\( t \\) is an integer such that \\( 8 t+1 \\) is a square, then \\( 8 t+1 \\) is the square of some odd integer \\( 2 n+1 \\), and hence \\( t=n(n+1) / 2 \\). Thus \\( t \\) is a triangular number if and only if \\( 8 t+1 \\) is a square. If \\( k \\) is an odd integer, then \\( k^{2} \\equiv 1(\\bmod 8) \\), and\n\\[\n\\begin{aligned}\nt \\text { is a triangular number } & \\Longleftrightarrow 8 t+1 \\text { is a square } \\\\\n& \\Longleftrightarrow k^{2}(8 t+1) \\text { is a square } \\\\\n& \\Longleftrightarrow 8\\left(k^{2} t+\\frac{k^{2}-1}{8}\\right)+1 \\text { is a square } \\\\\n& \\Longleftrightarrow\\left(k^{2} t+\\frac{k^{2}-1}{8}\\right) \\text { is a triangular number. }\n\\end{aligned}\n\\]\n\nHence we may take \\( (a, b)=\\left(k^{2},\\left(k^{2}-1\\right) / 8\\right) \\) for any odd integer \\( k \\).\nRemark. Let us call \\( (a, b) \\) a triangular pair if \\( a \\) and \\( b \\) are integers with the property that for positive integers \\( t, t \\) is a triangular number if and only if \\( a t+b \\) is a triangular number. Solution 2 showed that if \\( k \\) is an odd integer, then \\( \\left(k^{2},\\left(k^{2}-1\\right) / 8\\right) \\) is a triangular pair. In other words, the triangular pairs are of the form \\( \\left((2 m+1)^{2}, t_{m}\\right) \\), where \\( m \\) is any integer. We now show, conversely, that every triangular pair has this form.\n\nSuppose that \\( (a, b) \\) is a triangular pair. For any integer \\( n \\geq 0, a t_{n}+b \\) is triangular, so \\( 8\\left(a t_{n}+b\\right)+1=4 a n^{2}+4 a n+(8 b+1) \\) is a square. A polynomial \\( f(x) \\in \\mathbb{Z}[x] \\) taking square integer values at all nonnegative integers must be the square of a polynomial in \\( \\mathbb{Z}[x] \\) : see 1998B6 for a proof. Hence\n\\[\n4 a n^{2}+4 a n+(8 b+1)=\\ell(n)^{2}\n\\]\nfor some linear polynomial \\( \\ell(x) \\). Completing the square shows that \\( \\ell(x)=2 \\sqrt{a} x+\\sqrt{a} \\). Since \\( \\ell(n)^{2} \\) is the square of an integer for any integer \\( n, a=\\ell(0)^{2}=k^{2} \\) for some integer \\( k \\), and equating constant coefficients in (1) shows that \\( a=8 b+1 \\), so \\( b=\\left(k^{2}-1\\right) / 8 \\). Finally, \\( k \\) must be odd, in order for \\( b \\) to be an integer.",
  "vars": [
    "t",
    "t_n",
    "n",
    "x",
    "f",
    "\\\\ell"
  ],
  "params": [
    "a",
    "b",
    "k",
    "m"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "t": "timevar",
        "t_n": "triseq",
        "n": "indexer",
        "x": "varplace",
        "f": "polyfun",
        "\\ell": "linpoly",
        "a": "lincoef",
        "b": "shiftval",
        "k": "oddparam",
        "m": "baseint"
      },
      "question": "Prove that there exist an infinite number of ordered pairs $(lincoef,shiftval)$ of integers such that for every positive integer $timevar$, the number $lincoef\\,timevar+shiftval$ is a triangular number if and only if $timevar$ is a triangular number. (The triangular numbers are the $triseq = indexer(indexer+1)/2$ with $indexer$ in $\\{0,1,2,\\dots\\}$.)",
      "solution": "Solution 1. It is easy to check that \\( timevar_{3\\,indexer+1}=9\\,triseq+1 \\), while \\( timevar_{3\\,indexer} \\equiv timevar_{3\\,indexer+2} \\equiv 0 \\) \\((\\bmod 3)\\) for any integer \\( indexer \\). Hence for every positive integer \\( timevar, timevar \\) is a triangular number if and only if \\( 9\\,timevar+1 \\) is triangular. If the \\( indexer \\)th iterate of the linear map \\( timevar \\mapsto 9\\,timevar+1 \\) is \\( timevar \\mapsto lincoef\\,timevar+shiftval \\), then a chain of equivalences will show that \\( timevar \\) is a triangular number if and only if \\( lincoef\\,timevar+shiftval \\) is triangular. We obtain infinitely many pairs of integers \\( (lincoef,shiftval) \\) in this way.\n\nSolution 2. If \\( timevar=indexer(indexer+1)/2 \\), then \\( 8\\,timevar+1=(2\\,indexer+1)^{2} \\). Conversely, if \\( timevar \\) is an integer such that \\( 8\\,timevar+1 \\) is a square, then \\( 8\\,timevar+1 \\) is the square of some odd integer \\( 2\\,indexer+1 \\), and hence \\( timevar=indexer(indexer+1)/2 \\). Thus \\( timevar \\) is a triangular number if and only if \\( 8\\,timevar+1 \\) is a square. If \\( oddparam \\) is an odd integer, then \\( oddparam^{2} \\equiv 1\\,(\\bmod 8) \\), and\n\\[\n\\begin{aligned}\n timevar \\text{ is a triangular number } &\\Longleftrightarrow 8\\,timevar+1 \\text{ is a square } \\\\\n &\\Longleftrightarrow oddparam^{2}(8\\,timevar+1) \\text{ is a square } \\\\\n &\\Longleftrightarrow 8\\left(oddparam^{2}\\,timevar+\\frac{oddparam^{2}-1}{8}\\right)+1 \\text{ is a square } \\\\\n &\\Longleftrightarrow \\left(oddparam^{2}\\,timevar+\\frac{oddparam^{2}-1}{8}\\right) \\text{ is a triangular number. }\n\\end{aligned}\n\\]\nHence we may take \\( (lincoef,shiftval)=\\left(oddparam^{2},\\left(oddparam^{2}-1\\right)/8\\right) \\) for any odd integer \\( oddparam \\).\n\nRemark. Let us call \\( (lincoef,shiftval) \\) a triangular pair if \\( lincoef \\) and \\( shiftval \\) are integers with the property that for positive integers \\( timevar, timevar \\) is a triangular number if and only if \\( lincoef\\,timevar+shiftval \\) is a triangular number. Solution&nbsp;2 showed that if \\( oddparam \\) is an odd integer, then \\( \\left(oddparam^{2},\\left(oddparam^{2}-1\\right)/8\\right) \\) is a triangular pair. In other words, the triangular pairs are of the form \\( \\left((2\\,baseint+1)^{2},triseq\\right) \\), where \\( baseint \\) is any integer. We now show, conversely, that every triangular pair has this form.\n\nSuppose that \\( (lincoef,shiftval) \\) is a triangular pair. For any integer \\( indexer \\ge 0, lincoef\\,triseq+shiftval \\) is triangular, so \\( 8\\left(lincoef\\,triseq+shiftval\\right)+1=4\\,lincoef\\,indexer^{2}+4\\,lincoef\\,indexer+(8\\,shiftval+1) \\) is a square. A polynomial \\( polyfun(varplace) \\in \\mathbb{Z}[varplace] \\) taking square integer values at all non-negative integers must be the square of a polynomial in \\( \\mathbb{Z}[varplace] \\); see 1998B6 for a proof. Hence\n\\[\n4\\,lincoef\\,indexer^{2}+4\\,lincoef\\,indexer+(8\\,shiftval+1)=linpoly(indexer)^{2}\n\\]\nfor some linear polynomial \\( linpoly(varplace) \\). Completing the square shows that \\( linpoly(varplace)=2\\,\\sqrt{lincoef}\\,varplace+\\sqrt{lincoef} \\). Since \\( linpoly(indexer)^{2} \\) is the square of an integer for any integer \\( indexer \\), we have \\( lincoef=linpoly(0)^{2}=oddparam^{2} \\) for some integer \\( oddparam \\), and equating constant coefficients in (1) gives \\( lincoef=8\\,shiftval+1 \\), so \\( shiftval=\\left(oddparam^{2}-1\\right)/8 \\). Finally, \\( oddparam \\) must be odd in order for \\( shiftval \\) to be an integer."
    },
    "descriptive_long_confusing": {
      "map": {
        "t": "marigold",
        "t_n": "hummingbird",
        "n": "sandcastle",
        "x": "driftwood",
        "f": "sunflower",
        "\\ell": "gravelroad",
        "a": "watercolor",
        "b": "tangerine",
        "k": "toothbrush",
        "m": "firefly"
      },
      "question": "Prove that there exist an infinite number of ordered pairs $(watercolor,tangerine)$ of\nintegers such that for every positive integer $marigold$, the number $watercolormarigold+tangerine$\nis a triangular number if and only if $marigold$ is a triangular number. (The\ntriangular numbers are the $hummingbird = sandcastle(sandcastle+1)/2$ with $sandcastle$ in $\\{0,1,2,\\dots\\}$.)",
      "solution": "Solution 1. It is easy to check that \\( hummingbird_{3 sandcastle+1}=9 hummingbird_{sandcastle}+1 \\), while \\( hummingbird_{3 sandcastle} \\equiv hummingbird_{3 sandcastle+2} \\equiv 0 \\\\ (\\bmod 3) \\) for any integer \\( sandcastle \\). Hence for every positive integer \\( marigold, marigold \\) is a triangular number if and only if \\( 9 marigold+1 \\) is triangular. If the \\( sandcastle \\)th iterate of the linear map \\( marigold \\mapsto 9 marigold+1 \\) is \\( marigold \\mapsto watercolor marigold+tangerine \\), then a chain of equivalences will show that \\( marigold \\) is a triangular number if and only if \\( watercolor marigold+tangerine \\) is triangular. We obtain infinitely many pairs of integers \\( (watercolor, tangerine) \\) in this way.\n\nSolution 2. If \\( marigold=sandcastle(sandcastle+1) / 2 \\), then \\( 8 marigold+1=(2 sandcastle+1)^{2} \\). Conversely if \\( marigold \\) is an integer such that \\( 8 marigold+1 \\) is a square, then \\( 8 marigold+1 \\) is the square of some odd integer \\( 2 sandcastle+1 \\), and hence \\( marigold=sandcastle(sandcastle+1) / 2 \\). Thus \\( marigold \\) is a triangular number if and only if \\( 8 marigold+1 \\) is a square. If \\( toothbrush \\) is an odd integer, then \\( toothbrush^{2} \\equiv 1(\\bmod 8) \\), and\n\\[\n\\begin{aligned}\nmarigold \\text { is a triangular number } & \\Longleftrightarrow 8 marigold+1 \\text { is a square } \\\\\n& \\Longleftrightarrow toothbrush^{2}(8 marigold+1) \\text { is a square } \\\\\n& \\Longleftrightarrow 8\\left(toothbrush^{2} marigold+\\frac{toothbrush^{2}-1}{8}\\right)+1 \\text { is a square } \\\\\n& \\Longleftrightarrow\\left(toothbrush^{2} marigold+\\frac{toothbrush^{2}-1}{8}\\right) \\text { is a triangular number. }\n\\end{aligned}\n\\]\n\nHence we may take \\( (watercolor, tangerine)=\\left(toothbrush^{2},\\left(toothbrush^{2}-1\\right) / 8\\right) \\) for any odd integer \\( toothbrush \\).\n\nRemark. Let us call \\( (watercolor, tangerine) \\) a triangular pair if \\( watercolor \\) and \\( tangerine \\) are integers with the property that for positive integers \\( marigold, marigold \\) is a triangular number if and only if \\( watercolor marigold+tangerine \\) is a triangular number. Solution 2 showed that if \\( toothbrush \\) is an odd integer, then \\( \\left(toothbrush^{2},\\left(toothbrush^{2}-1\\right) / 8\\right) \\) is a triangular pair. In other words, the triangular pairs are of the form \\( \\left((2 firefly+1)^{2}, hummingbird_{firefly}\\right) \\), where \\( firefly \\) is any integer. We now show, conversely, that every triangular pair has this form.\n\nSuppose that \\( (watercolor, tangerine) \\) is a triangular pair. For any integer \\( sandcastle \\geq 0, watercolor hummingbird_{sandcastle}+tangerine \\) is triangular, so \\( 8\\left(watercolor hummingbird_{sandcastle}+tangerine\\right)+1=4 watercolor sandcastle^{2}+4 watercolor sandcastle+(8 tangerine+1) \\) is a square. A polynomial \\( sunflower(driftwood) \\in \\mathbb{Z}[driftwood] \\) taking square integer values at all nonnegative integers must be the square of a polynomial in \\( \\mathbb{Z}[driftwood] \\) : see 1998B6 for a proof. Hence\n\\[\n4 watercolor sandcastle^{2}+4 watercolor sandcastle+(8 tangerine+1)=gravelroad(sandcastle)^{2}\n\\]\nfor some linear polynomial \\( gravelroad(driftwood) \\). Completing the square shows that \\( gravelroad(driftwood)=2 \\sqrt{watercolor} driftwood+\\sqrt{watercolor} \\). Since \\( gravelroad(sandcastle)^{2} \\) is the square of an integer for any integer \\( sandcastle, watercolor=gravelroad(0)^{2}=toothbrush^{2} \\) for some integer \\( toothbrush \\), and equating constant coefficients in (1) shows that \\( watercolor=8 tangerine+1 \\), so \\( tangerine=\\left(toothbrush^{2}-1\\right) / 8 \\). Finally, \\( toothbrush \\) must be odd, in order for \\( tangerine \\) to be an integer."
    },
    "descriptive_long_misleading": {
      "map": {
        "t": "circularvalue",
        "t_n": "circularseqterm",
        "n": "continuousindex",
        "x": "dependentvalue",
        "f": "constantthing",
        "\\ell": "nonlinearpoly",
        "a": "divisorparam",
        "b": "multiplierparam",
        "k": "eveninteger",
        "m": "fractionalnum"
      },
      "question": "Prove that there exist an infinite number of ordered pairs $(divisorparam,multiplierparam)$ of\nintegers such that for every positive integer $circularvalue$, the number $divisorparam circularvalue+multiplierparam$\nis a triangular number if and only if $circularvalue$ is a triangular number. (The\ntriangular numbers are the $circularseqterm = continuousindex(continuousindex+1)/2$ with $continuousindex$ in $\\{0,1,2,\\dots\\}$.)",
      "solution": "Solution 1. It is easy to check that \\( circularvalue_{3 continuousindex+1}=9 circularvalue_{continuousindex}+1 \\), while \\( circularvalue_{3 continuousindex} \\equiv circularvalue_{3 continuousindex+2} \\equiv 0 \\) \\( (\\bmod 3) \\) for any integer \\( continuousindex \\). Hence for every positive integer \\( circularvalue, circularvalue \\) is a triangular number if and only if \\( 9 circularvalue+1 \\) is triangular. If the \\( continuousindex \\)th iterate of the linear map \\( circularvalue \\mapsto 9 circularvalue+1 \\) is \\( circularvalue \\mapsto divisorparam circularvalue+multiplierparam \\), then a chain of equivalences will show that \\( circularvalue \\) is a triangular number if and only if \\( divisorparam circularvalue+multiplierparam \\) is triangular. We obtain infinitely many pairs of integers \\( (divisorparam, multiplierparam) \\) in this way.\n\nSolution 2. If \\( circularvalue=continuousindex(continuousindex+1) / 2 \\), then \\( 8 circularvalue+1=(2 continuousindex+1)^{2} \\). Conversely if \\( circularvalue \\) is an integer such that \\( 8 circularvalue+1 \\) is a square, then \\( 8 circularvalue+1 \\) is the square of some odd integer \\( 2 continuousindex+1 \\), and hence \\( circularvalue=continuousindex(continuousindex+1) / 2 \\). Thus \\( circularvalue \\) is a triangular number if and only if \\( 8 circularvalue+1 \\) is a square. If \\( eveninteger \\) is an odd integer, then \\( eveninteger^{2} \\equiv 1(\\bmod 8) \\), and\n\\[\n\\begin{aligned}\ncircularvalue \\text { is a triangular number } & \\Longleftrightarrow 8 circularvalue+1 \\text { is a square } \\\\\n& \\Longleftrightarrow eveninteger^{2}(8 circularvalue+1) \\text { is a square } \\\\\n& \\Longleftrightarrow 8\\left(eveninteger^{2} circularvalue+\\frac{eveninteger^{2}-1}{8}\\right)+1 \\text { is a square } \\\\\n& \\Longleftrightarrow\\left(eveninteger^{2} circularvalue+\\frac{eveninteger^{2}-1}{8}\\right) \\text { is a triangular number. }\n\\end{aligned}\n\\]\n\nHence we may take \\( (divisorparam, multiplierparam)=\\left(eveninteger^{2},\\left(eveninteger^{2}-1\\right) / 8\\right) \\) for any odd integer \\( eveninteger \\).\nRemark. Let us call \\( (divisorparam, multiplierparam) \\) a triangular pair if \\( divisorparam \\) and \\( multiplierparam \\) are integers with the property that for positive integers \\( circularvalue, circularvalue \\) is a triangular number if and only if \\( divisorparam circularvalue+multiplierparam \\) is a triangular number. Solution 2 showed that if \\( eveninteger \\) is an odd integer, then \\( \\left(eveninteger^{2},\\left(eveninteger^{2}-1\\right) / 8\\right) \\) is a triangular pair. In other words, the triangular pairs are of the form \\( \\left((2 fractionalnum+1)^{2}, circularvalue_{fractionalnum}\\right) \\), where \\( fractionalnum \\) is any integer. We now show, conversely, that every triangular pair has this form.\n\nSuppose that \\( (divisorparam, multiplierparam) \\) is a triangular pair. For any integer \\( continuousindex \\geq 0, divisorparam circularseqterm_{continuousindex}+multiplierparam \\) is triangular, so \\( 8\\left(divisorparam circularseqterm_{continuousindex}+multiplierparam\\right)+1=4 divisorparam continuousindex^{2}+4 divisorparam continuousindex+(8 multiplierparam+1) \\) is a square. A polynomial \\( constantthing(dependentvalue) \\in \\mathbb{Z}[dependentvalue] \\) taking square integer values at all nonnegative integers must be the square of a polynomial in \\( \\mathbb{Z}[dependentvalue] \\) : see 1998B6 for a proof. Hence\n\\[\n4 divisorparam continuousindex^{2}+4 divisorparam continuousindex+(8 multiplierparam+1)=nonlinearpoly(continuousindex)^{2}\n\\]\nfor some linear polynomial \\( nonlinearpoly(dependentvalue) \\). Completing the square shows that \\( nonlinearpoly(dependentvalue)=2 \\sqrt{divisorparam} dependentvalue+\\sqrt{divisorparam} \\). Since \\( nonlinearpoly(continuousindex)^{2} \\) is the square of an integer for any integer \\( continuousindex, divisorparam=nonlinearpoly(0)^{2}=eveninteger^{2} \\) for some integer \\( eveninteger \\), and equating constant coefficients in (1) shows that \\( divisorparam=8 multiplierparam+1 \\), so \\( multiplierparam=\\left(eveninteger^{2}-1\\right) / 8 \\). Finally, \\( eveninteger \\) must be odd, in order for \\( multiplierparam \\) to be an integer."
    },
    "garbled_string": {
      "map": {
        "t": "qzxwvtnp",
        "t_n": "hjgrksla",
        "n": "vrclpzke",
        "x": "dmqsfyra",
        "f": "knbthwue",
        "\\ell": "pzgxmhio",
        "a": "wsnevuda",
        "b": "lqtrzako",
        "k": "gdimfscu",
        "m": "sojhapkr"
      },
      "question": "Prove that there exist an infinite number of ordered pairs $(wsnevuda,lqtrzako)$ of\nintegers such that for every positive integer $qzxwvtnp$, the number $wsnevuda qzxwvtnp+lqtrzako$\nis a triangular number if and only if $qzxwvtnp$ is a triangular number. (The\ntriangular numbers are the $hjgrksla = vrclpzke(vrclpzke+1)/2$ with $vrclpzke$ in $\\{0,1,2,\\dots\\}$.)\n\n\\end{itemize}\n\n\\end{document}",
      "solution": "Solution 1. It is easy to check that \\( qzxwvtnp_{3\\,vrclpzke+1}=9\\,qzxwvtnp_{vrclpzke}+1 \\), while \\( qzxwvtnp_{3\\,vrclpzke} \\equiv qzxwvtnp_{3\\,vrclpzke+2} \\equiv 0 \\)(\\bmod 3) for any integer \\( vrclpzke \\). Hence for every positive integer \\( qzxwvtnp, qzxwvtnp \\) is a triangular number if and only if \\( 9\\,qzxwvtnp+1 \\) is triangular. If the \\( vrclpzke \\)th iterate of the linear map \\( qzxwvtnp \\mapsto 9\\,qzxwvtnp+1 \\) is \\( qzxwvtnp \\mapsto wsnevuda\\,qzxwvtnp+lqtrzako \\), then a chain of equivalences will show that \\( qzxwvtnp \\) is a triangular number if and only if \\( wsnevuda\\,qzxwvtnp+lqtrzako \\) is triangular. We obtain infinitely many pairs of integers \\( (wsnevuda,lqtrzako) \\) in this way.\n\nSolution 2. If \\( qzxwvtnp=vrclpzke(vrclpzke+1)/2 \\), then \\( 8\\,qzxwvtnp+1=(2\\,vrclpzke+1)^{2} \\). Conversely if \\( qzxwvtnp \\) is an integer such that \\( 8\\,qzxwvtnp+1 \\) is a square, then \\( 8\\,qzxwvtnp+1 \\) is the square of some odd integer \\( 2\\,vrclpzke+1 \\), and hence \\( qzxwvtnp=vrclpzke(vrclpzke+1)/2 \\). Thus \\( qzxwvtnp \\) is a triangular number if and only if \\( 8\\,qzxwvtnp+1 \\) is a square. If \\( gdimfscu \\) is an odd integer, then \\( gdimfscu^{2} \\equiv 1(\\bmod 8) \\), and\n\\[\n\\begin{aligned}\nqzxwvtnp \\text { is a triangular number } & \\Longleftrightarrow 8\\,qzxwvtnp+1 \\text { is a square } \\\\\n& \\Longleftrightarrow gdimfscu^{2}(8\\,qzxwvtnp+1) \\text { is a square } \\\\\n& \\Longleftrightarrow 8\\left(gdimfscu^{2}\\,qzxwvtnp+\\frac{gdimfscu^{2}-1}{8}\\right)+1 \\text { is a square } \\\\\n& \\Longleftrightarrow\\left(gdimfscu^{2}\\,qzxwvtnp+\\frac{gdimfscu^{2}-1}{8}\\right) \\text { is a triangular number. }\n\\end{aligned}\n\\]\n\nHence we may take \\( (wsnevuda,lqtrzako)=\\left(gdimfscu^{2},\\left(gdimfscu^{2}-1\\right) / 8\\right) \\) for any odd integer \\( gdimfscu \\).\n\nRemark. Let us call \\( (wsnevuda,lqtrzako) \\) a triangular pair if \\( wsnevuda \\) and \\( lqtrzako \\) are integers with the property that for positive integers \\( qzxwvtnp, qzxwvtnp \\) is a triangular number if and only if \\( wsnevuda\\,qzxwvtnp+lqtrzako \\) is a triangular number. Solution 2 showed that if \\( gdimfscu \\) is an odd integer, then \\( \\left(gdimfscu^{2},\\left(gdimfscu^{2}-1\\right) / 8\\right) \\) is a triangular pair. In other words, the triangular pairs are of the form \\( \\left((2\\,sojhapkr+1)^{2}, qzxwvtnp_{sojhapkr}\\right) \\), where \\( sojhapkr \\) is any integer. We now show, conversely, that every triangular pair has this form.\n\nSuppose that \\( (wsnevuda,lqtrzako) \\) is a triangular pair. For any integer \\( vrclpzke \\geq 0, wsnevuda\\,qzxwvtnp_{vrclpzke}+lqtrzako \\) is triangular, so \\( 8\\left(wsnevuda\\,qzxwvtnp_{vrclpzke}+lqtrzako\\right)+1=4\\,wsnevuda\\,vrclpzke^{2}+4\\,wsnevuda\\,vrclpzke+(8\\,lqtrzako+1) \\) is a square. A polynomial \\( knbthwue(dmqsfyra) \\in \\mathbb{Z}[dmqsfyra] \\) taking square integer values at all nonnegative integers must be the square of a polynomial in \\( \\mathbb{Z}[dmqsfyra] \\) : see 1998B6 for a proof. Hence\n\\[\n4\\,wsnevuda\\,vrclpzke^{2}+4\\,wsnevuda\\,vrclpzke+(8\\,lqtrzako+1)=pzgxmhio(vrclpzke)^{2}\n\\]\nfor some linear polynomial \\( pzgxmhio(dmqsfyra) \\). Completing the square shows that \\( pzgxmhio(dmqsfyra)=2 \\sqrt{wsnevuda}\\,dmqsfyra+\\sqrt{wsnevuda} \\). Since \\( pzgxmhio(vrclpzke)^{2} \\) is the square of an integer for any integer \\( vrclpzke, wsnevuda=pzgxmhio(0)^{2}=gdimfscu^{2} \\) for some integer \\( gdimfscu \\), and equating constant coefficients in (1) shows that \\( wsnevuda=8\\,lqtrzako+1 \\), so \\( lqtrzako=\\left(gdimfscu^{2}-1\\right) / 8 \\). Finally, \\( gdimfscu \\) must be odd, in order for \\( lqtrzako \\) to be an integer."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nT=\\bigl\\{\\tau_{n}=n(n+1)/2 : n\\ge 0\\bigr\\}, \\qquad  \nP=\\bigl\\{\\pi_{n}=n(3n-1)/2 : n\\ge 0\\bigr\\},\n\\]\nand for integers $A,B,C$ put  \n\\[\nF_{A,B}(t)=At+B,\\qquad G_{A,C}(t)=At+C,\\qquad t\\in\\mathbb Z_{\\ge 1}.\n\\]\n\nProve that there exist infinitely many triples $(A,B,C)\\in\\mathbb Z^{3}$ that satisfy  \n\n(i) $A=p^{2}$ where $p$ is a prime with $p\\equiv 1\\pmod 6$;  \n\n(ii) for every positive integer $t$\n\\[\nt\\in T\\;\\Longleftrightarrow\\;F_{A,B}(t)\\in T,\n\\qquad\nt\\in P\\;\\Longleftrightarrow\\;G_{A,C}(t)\\in P .\n\\]\n\nMoreover, determine all such triples $(A,B,C)$.\n\n(Thus a common dilation factor equal to the square of a prime $p\\equiv 1\\pmod 6$ permutes the triangular numbers after the translation $B$ and the pentagonal numbers after the translation $C$.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "All congruences are taken modulo the stated modulus.\n\nStep 0.  Characteristic tests for the two polygonal sequences.  \n(a)  For every $t\\ge 0$\n\\[\nt\\in T\\Longleftrightarrow 8t+1 \\text{ is a perfect square.}\n\\]  \n(b)  For every $t>0$\n\\[\nt\\in P \\Longleftrightarrow \\exists\\,s\\in\\mathbb Z_{>0}:\\;\n        s^{2}=24t+1,\\;s\\equiv 5\\pmod 6.\n\\]\n(When $t=0$ one has $24t+1=1^{2}$ with the unique positive root $s=1\\equiv 1\\pmod 6$; we never need this special case, because the whole problem is formulated for $t\\ge 1$.)\n\nProof of (b).  \nIf $t=\\pi_{n}$ with $n\\ge 1$, then $24t+1=(6n-1)^{2}$ and the root $6n-1$ is positive and congruent to $5$ modulo $6$.  \nConversely, if $t>0$ and $s^{2}=24t+1$ with $s\\equiv 5\\pmod 6$, then $s=6n-1$ for some $n\\ge 1$, hence $t=\\pi_{n}$.\n\nA classical device used repeatedly below is the  \n\nPutnam–B6 lemma.  \nLet $f(x)\\in\\mathbb Z[x]$ be quadratic.  If $f(n)$ is a perfect square for infinitely many integers $n$, then $f(x)$ is the square of a linear polynomial in $\\mathbb Z[x]$.\n\n---------------------------------------------------------------\nPart I.  Affine self-maps of the triangular numbers  \n\nLemma 1.  \nA pair $(a,b)\\in\\mathbb Z^{2}$ satisfies  \n\\[\nt\\in T\\;\\Longleftrightarrow\\;at+b\\in T\n\\quad(\\forall\\,t\\ge 1)\n\\]\niff there is an odd integer $k$ such that  \n\\[\na=k^{2},\\qquad b=\\dfrac{k^{2}-1}{8}.\n\\]\n\nProof.  Substituting $t=\\tau_{n}$ with $n\\ge 1$ gives  \n\\[\n8\\bigl(a\\tau_{n}+b\\bigr)+1=4an^{2}+4an+(8b+1)\n\\]\nsquare for infinitely many $n$.  The Putnam–B6 lemma yields  \n\\[\n4an^{2}+4an+(8b+1)=\\bigl(cn+d\\bigr)^{2}\\quad(c,d\\in\\mathbb Z).\n\\]\nComparing coefficients gives $4a=c^{2}\\,(=4k^{2})$ and $8b+1=d^{2}=k^{2}$, i.e. the asserted formulas.  \nConversely,\n\\[\n8\\bigl(at+b\\bigr)+1=k^{2}(8t+1)\n\\]\nis a square $\\Longleftrightarrow 8t+1$ is a square, proving the equivalence. \\qed\n\n----------------------------------------------------------------\nPart II.  Affine self-maps of the pentagonal numbers  \n\nLemma 2.  \nA pair $(a,c)\\in\\mathbb Z^{2}$ satisfies  \n\\[\nt\\in P\\;\\Longleftrightarrow\\;at+c\\in P\n\\quad(\\forall\\,t\\ge 1)\n\\]\niff there exists an integer $r$ such that  \n\\[\na=r^{2},\\qquad c=\\dfrac{r^{2}-1}{24},\\qquad r\\equiv 1\\pmod 6.\n\\]\n\nProof (necessity).  \nInsert $t=\\pi_{n}$ with $n\\ge 1$.  Using Step 0 we obtain  \n\\[\n24\\bigl(a\\pi_{n}+c\\bigr)+1=36an^{2}-12an+(24c+1)\n\\]\nsquare for infinitely many $n$.  Putnam–B6 gives  \n\\[\n36an^{2}-12an+24c+1=(en+f)^{2}\\quad(e,f\\in\\mathbb Z).\n\\]\nMatching coefficients yields  \n\\[\ne=6r,\\;a=r^{2};\\qquad f=-r;\\qquad r^{2}=24c+1.\n\\]\nSo far we have $a=r^{2}$ and $c=(r^{2}-1)/24$.\n\nWe still have to prove $r\\equiv 1\\pmod 6$.  \nTake $t=\\pi_{1}=1\\in P$.  Then\n\\[\n24(at+c)+1=r^{2}\\bigl(24\\cdot 1+1\\bigr)=25r^{2}=(5r)^{2}.\n\\]\nSince $at+c\\in P$, the pentagonal test forces $5r\\equiv 5\\pmod 6$, hence $r\\equiv 1\\pmod 6$.  This finishes the necessity part.\n\nProof (sufficiency).  \nAssume $a=r^{2}$, $c=(r^{2}-1)/24$, $r\\equiv 1\\pmod 6$.  \nFor $t\\in P$, pick $s>0$ with $s^{2}=24t+1$ and $s\\equiv 5\\pmod 6$.  Then  \n\\[\n24\\bigl(at+c\\bigr)+1=r^{2}\\bigl(24t+1\\bigr)=(rs)^{2},\n\\qquad rs\\equiv 1\\cdot 5\\equiv 5\\pmod 6,\n\\]\nso $at+c\\in P$.\n\nConversely suppose $at+c\\in P$.  Hence  \n\\[\ns'^{2}=24\\bigl(at+c\\bigr)+1=r^{2}\\bigl(24t+1\\bigr)\n\\]\nfor some $s'\\equiv 5\\pmod 6$.  Let $q$ be any prime divisor of $r$.  \nFrom $s'^{2}=r^{2}(24t+1)$ we get  \n\\[\nv_{q}\\bigl(s'^{2}\\bigr)=2v_{q}(s')=2v_{q}(r)+v_{q}\\bigl(24t+1\\bigr),\n\\]\nhence $v_{q}(s')\\ge v_{q}(r)$ and $q\\mid s'$.  Doing this for every $q\\mid r$ shows $r\\mid s'$, so write $s'=rs$ with $s\\in\\mathbb Z_{>0}$.  Then  \n\\[\ns^{2}=24t+1,\\qquad s\\equiv \\dfrac{s'}{r}\\equiv\\dfrac{5}{1}\\equiv 5\\pmod 6,\n\\]\nwhence $t\\in P$.  Thus the equivalence holds. \\qed\n\n----------------------------------------------------------------\nPart III.  Triples working for both polygonal families  \n\nIf $(A,B,C)$ satisfies (ii), Lemmas 1–2 give integers $k,r$ with  \n\\[\nA=k^{2}=r^{2},\\quad\nB=\\dfrac{k^{2}-1}{8},\\quad\nC=\\dfrac{r^{2}-1}{24},\\quad\nk\\ \\text{odd},\\;r\\equiv 1\\pmod 6.\n\\]\nHence $k=\\pm r$, and changing the sign of $k$ leaves $B,C$ unchanged; we may assume $k=r$.  Writing $k$ instead of $r$ we obtain  \n\\[\nA=k^{2},\\qquad B=\\dfrac{k^{2}-1}{8},\\qquad C=\\dfrac{k^{2}-1}{24},\n\\qquad k\\ \\text{odd},\\;k\\equiv 1\\pmod 6.\n\\tag{1}\n\\]\n\nConversely, every odd $k\\equiv 1\\pmod 6$ gives a triple in (1) satisfying (ii).\n\n----------------------------------------------------------------\nPart IV.  Imposing the prime-square restriction  \n\nCondition (i) forces $A=p^{2}$ with $p$ prime, $p\\equiv 1\\pmod 6$.  Putting $k=p$ in (1) yields exactly  \n\\[\nA=p^{2},\\qquad B=\\dfrac{p^{2}-1}{8},\\qquad C=\\dfrac{p^{2}-1}{24},\n\\qquad p\\ \\text{prime},\\;p\\equiv 1\\pmod 6.\n\\tag{2}\n\\]\nSince $p^{2}\\equiv 1\\pmod{24}$, both fractions are integers.  Dirichlet’s theorem provides infinitely many such primes, hence infinitely many admissible triples.  Finally, if $(A,B,C)$ satisfies (ii) and $A=p^{2}$ with $p$ prime, then $k=p$ in (1); therefore $(B,C)$ must coincide with the values in (2).  No other triples exist.\n\n----------------------------------------------------------------\nConclusion.  \nAll ordered triples required in the problem are  \n\\[\n\\boxed{\\;\n  \\Bigl\\{\n     \\bigl(p^{2},\\,\\tfrac{p^{2}-1}{8},\\,\\tfrac{p^{2}-1}{24}\\bigr)\n        :\\; p\\ \\text{prime},\\;p\\equiv 1\\pmod 6\n  \\Bigr\\}\\;}\n\\]\nand there are infinitely many of them. \\qed\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.707609",
        "was_fixed": false,
        "difficulty_analysis": "1. Two interacting structures.  The problem now demands simultaneous control of both triangular and pentagonal numbers, forcing a delicate intersection of two quadratic-form characterisations (8t+1 square, 24t+1 square ≡ 5 mod 6).  \n\n2. Classification twice.  One must re-prove the “square-polynomial” rigidity for pentagonal numbers, an argument absent from the original task.  Handling the extra mod 6 congruence raises the algebraic subtlety.  \n\n3. Compatibility constraint.  After the separate classifications the solver still has to reconcile them, solving k² = r² and congruence conditions, then verifying integrality of the two distinct translations B and C.  \n\n4. Arithmetic of primes in progressions.  To exhibit infinitely many examples with A a square of an odd prime (rather than of an arbitrary odd integer) the solution invokes Dirichlet’s theorem – a result well beyond elementary number theory.  \n\n5. Final characterisation.  Unlike the original problem, which only shows existence, the enhanced variant requires a full description of all admissible triples, raising the bar from construction to complete classification."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nT=\\bigl\\{\\tau_{n}=n(n+1)/2 : n\\ge 0\\bigr\\}, \\qquad  \nP=\\bigl\\{\\pi_{n}=n(3n-1)/2 : n\\ge 0\\bigr\\},\n\\]\nand for integers $A,B,C$ put  \n\\[\nF_{A,B}(t)=At+B,\\qquad G_{A,C}(t)=At+C,\\qquad t\\in\\mathbb Z_{\\ge 1}.\n\\]\n\nProve that there exist infinitely many triples $(A,B,C)\\in\\mathbb Z^{3}$ that satisfy  \n\n(i) $A=p^{2}$ where $p$ is a prime with $p\\equiv 1\\pmod 6$;  \n\n(ii) for every positive integer $t$\n\\[\nt\\in T\\;\\Longleftrightarrow\\;F_{A,B}(t)\\in T,\n\\qquad\nt\\in P\\;\\Longleftrightarrow\\;G_{A,C}(t)\\in P .\n\\]\n\nMoreover, determine all such triples $(A,B,C)$.\n\n(Thus a common dilation factor equal to the square of a prime $p\\equiv 1\\pmod 6$ permutes the triangular numbers after the translation $B$ and the pentagonal numbers after the translation $C$.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "All congruences are taken modulo the stated modulus.\n\nStep 0.  Characteristic tests for the two polygonal sequences.  \n(a)  For every $t\\ge 0$\n\\[\nt\\in T\\Longleftrightarrow 8t+1 \\text{ is a perfect square.}\n\\]  \n(b)  For every $t>0$\n\\[\nt\\in P \\Longleftrightarrow \\exists\\,s\\in\\mathbb Z_{>0}:\\;\n        s^{2}=24t+1,\\;s\\equiv 5\\pmod 6.\n\\]\n(When $t=0$ one has $24t+1=1^{2}$ with the unique positive root $s=1\\equiv 1\\pmod 6$; we never need this special case, because the whole problem is formulated for $t\\ge 1$.)\n\nProof of (b).  \nIf $t=\\pi_{n}$ with $n\\ge 1$, then $24t+1=(6n-1)^{2}$ and the root $6n-1$ is positive and congruent to $5$ modulo $6$.  \nConversely, if $t>0$ and $s^{2}=24t+1$ with $s\\equiv 5\\pmod 6$, then $s=6n-1$ for some $n\\ge 1$, hence $t=\\pi_{n}$.\n\nA classical device used repeatedly below is the  \n\nPutnam–B6 lemma.  \nLet $f(x)\\in\\mathbb Z[x]$ be quadratic.  If $f(n)$ is a perfect square for infinitely many integers $n$, then $f(x)$ is the square of a linear polynomial in $\\mathbb Z[x]$.\n\n---------------------------------------------------------------\nPart I.  Affine self-maps of the triangular numbers  \n\nLemma 1.  \nA pair $(a,b)\\in\\mathbb Z^{2}$ satisfies  \n\\[\nt\\in T\\;\\Longleftrightarrow\\;at+b\\in T\n\\quad(\\forall\\,t\\ge 1)\n\\]\niff there is an odd integer $k$ such that  \n\\[\na=k^{2},\\qquad b=\\dfrac{k^{2}-1}{8}.\n\\]\n\nProof.  Substituting $t=\\tau_{n}$ with $n\\ge 1$ gives  \n\\[\n8\\bigl(a\\tau_{n}+b\\bigr)+1=4an^{2}+4an+(8b+1)\n\\]\nsquare for infinitely many $n$.  The Putnam–B6 lemma yields  \n\\[\n4an^{2}+4an+(8b+1)=\\bigl(cn+d\\bigr)^{2}\\quad(c,d\\in\\mathbb Z).\n\\]\nComparing coefficients gives $4a=c^{2}\\,(=4k^{2})$ and $8b+1=d^{2}=k^{2}$, i.e. the asserted formulas.  \nConversely,\n\\[\n8\\bigl(at+b\\bigr)+1=k^{2}(8t+1)\n\\]\nis a square $\\Longleftrightarrow 8t+1$ is a square, proving the equivalence. \\qed\n\n----------------------------------------------------------------\nPart II.  Affine self-maps of the pentagonal numbers  \n\nLemma 2.  \nA pair $(a,c)\\in\\mathbb Z^{2}$ satisfies  \n\\[\nt\\in P\\;\\Longleftrightarrow\\;at+c\\in P\n\\quad(\\forall\\,t\\ge 1)\n\\]\niff there exists an integer $r$ such that  \n\\[\na=r^{2},\\qquad c=\\dfrac{r^{2}-1}{24},\\qquad r\\equiv 1\\pmod 6.\n\\]\n\nProof (necessity).  \nInsert $t=\\pi_{n}$ with $n\\ge 1$.  Using Step 0 we obtain  \n\\[\n24\\bigl(a\\pi_{n}+c\\bigr)+1=36an^{2}-12an+(24c+1)\n\\]\nsquare for infinitely many $n$.  Putnam–B6 gives  \n\\[\n36an^{2}-12an+24c+1=(en+f)^{2}\\quad(e,f\\in\\mathbb Z).\n\\]\nMatching coefficients yields  \n\\[\ne=6r,\\;a=r^{2};\\qquad f=-r;\\qquad r^{2}=24c+1.\n\\]\nSo far we have $a=r^{2}$ and $c=(r^{2}-1)/24$.\n\nWe still have to prove $r\\equiv 1\\pmod 6$.  \nTake $t=\\pi_{1}=1\\in P$.  Then\n\\[\n24(at+c)+1=r^{2}\\bigl(24\\cdot 1+1\\bigr)=25r^{2}=(5r)^{2}.\n\\]\nSince $at+c\\in P$, the pentagonal test forces $5r\\equiv 5\\pmod 6$, hence $r\\equiv 1\\pmod 6$.  This finishes the necessity part.\n\nProof (sufficiency).  \nAssume $a=r^{2}$, $c=(r^{2}-1)/24$, $r\\equiv 1\\pmod 6$.  \nFor $t\\in P$, pick $s>0$ with $s^{2}=24t+1$ and $s\\equiv 5\\pmod 6$.  Then  \n\\[\n24\\bigl(at+c\\bigr)+1=r^{2}\\bigl(24t+1\\bigr)=(rs)^{2},\n\\qquad rs\\equiv 1\\cdot 5\\equiv 5\\pmod 6,\n\\]\nso $at+c\\in P$.\n\nConversely suppose $at+c\\in P$.  Hence  \n\\[\ns'^{2}=24\\bigl(at+c\\bigr)+1=r^{2}\\bigl(24t+1\\bigr)\n\\]\nfor some $s'\\equiv 5\\pmod 6$.  Let $q$ be any prime divisor of $r$.  \nFrom $s'^{2}=r^{2}(24t+1)$ we get  \n\\[\nv_{q}\\bigl(s'^{2}\\bigr)=2v_{q}(s')=2v_{q}(r)+v_{q}\\bigl(24t+1\\bigr),\n\\]\nhence $v_{q}(s')\\ge v_{q}(r)$ and $q\\mid s'$.  Doing this for every $q\\mid r$ shows $r\\mid s'$, so write $s'=rs$ with $s\\in\\mathbb Z_{>0}$.  Then  \n\\[\ns^{2}=24t+1,\\qquad s\\equiv \\dfrac{s'}{r}\\equiv\\dfrac{5}{1}\\equiv 5\\pmod 6,\n\\]\nwhence $t\\in P$.  Thus the equivalence holds. \\qed\n\n----------------------------------------------------------------\nPart III.  Triples working for both polygonal families  \n\nIf $(A,B,C)$ satisfies (ii), Lemmas 1–2 give integers $k,r$ with  \n\\[\nA=k^{2}=r^{2},\\quad\nB=\\dfrac{k^{2}-1}{8},\\quad\nC=\\dfrac{r^{2}-1}{24},\\quad\nk\\ \\text{odd},\\;r\\equiv 1\\pmod 6.\n\\]\nHence $k=\\pm r$, and changing the sign of $k$ leaves $B,C$ unchanged; we may assume $k=r$.  Writing $k$ instead of $r$ we obtain  \n\\[\nA=k^{2},\\qquad B=\\dfrac{k^{2}-1}{8},\\qquad C=\\dfrac{k^{2}-1}{24},\n\\qquad k\\ \\text{odd},\\;k\\equiv 1\\pmod 6.\n\\tag{1}\n\\]\n\nConversely, every odd $k\\equiv 1\\pmod 6$ gives a triple in (1) satisfying (ii).\n\n----------------------------------------------------------------\nPart IV.  Imposing the prime-square restriction  \n\nCondition (i) forces $A=p^{2}$ with $p$ prime, $p\\equiv 1\\pmod 6$.  Putting $k=p$ in (1) yields exactly  \n\\[\nA=p^{2},\\qquad B=\\dfrac{p^{2}-1}{8},\\qquad C=\\dfrac{p^{2}-1}{24},\n\\qquad p\\ \\text{prime},\\;p\\equiv 1\\pmod 6.\n\\tag{2}\n\\]\nSince $p^{2}\\equiv 1\\pmod{24}$, both fractions are integers.  Dirichlet’s theorem provides infinitely many such primes, hence infinitely many admissible triples.  Finally, if $(A,B,C)$ satisfies (ii) and $A=p^{2}$ with $p$ prime, then $k=p$ in (1); therefore $(B,C)$ must coincide with the values in (2).  No other triples exist.\n\n----------------------------------------------------------------\nConclusion.  \nAll ordered triples required in the problem are  \n\\[\n\\boxed{\\;\n  \\Bigl\\{\n     \\bigl(p^{2},\\,\\tfrac{p^{2}-1}{8},\\,\\tfrac{p^{2}-1}{24}\\bigr)\n        :\\; p\\ \\text{prime},\\;p\\equiv 1\\pmod 6\n  \\Bigr\\}\\;}\n\\]\nand there are infinitely many of them. \\qed\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.552202",
        "was_fixed": false,
        "difficulty_analysis": "1. Two interacting structures.  The problem now demands simultaneous control of both triangular and pentagonal numbers, forcing a delicate intersection of two quadratic-form characterisations (8t+1 square, 24t+1 square ≡ 5 mod 6).  \n\n2. Classification twice.  One must re-prove the “square-polynomial” rigidity for pentagonal numbers, an argument absent from the original task.  Handling the extra mod 6 congruence raises the algebraic subtlety.  \n\n3. Compatibility constraint.  After the separate classifications the solver still has to reconcile them, solving k² = r² and congruence conditions, then verifying integrality of the two distinct translations B and C.  \n\n4. Arithmetic of primes in progressions.  To exhibit infinitely many examples with A a square of an odd prime (rather than of an arbitrary odd integer) the solution invokes Dirichlet’s theorem – a result well beyond elementary number theory.  \n\n5. Final characterisation.  Unlike the original problem, which only shows existence, the enhanced variant requires a full description of all admissible triples, raising the bar from construction to complete classification."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}