{
  "index": "1960-B-4",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "4. Consider the arithmetic progression \\( a, a+d, a+2 d, \\ldots \\), where \\( a \\) and \\( d \\) are positive integers. For any positive integer \\( k \\), prove that the progression has either no exact \\( k \\) th powers or infinitely many.",
  "solution": "Solution. Suppose the given arithmetic progression contains \\( n^{k} \\) (i.e., it contains a perfect \\( k \\) th power). Then the equation\n\\[\n(n+d)^{k}=n^{k}+d\\left|\\binom{k}{1} n^{k} 1+\\binom{k}{2} n^{k} 2 d+\\cdots+d^{k} 1\\right|\n\\]\nshows that \\( (n+d)^{k} \\) is in the progression. By induction, \\( (n+2 d)^{k} \\), \\( (n+3 d)^{k}, \\ldots \\) are also in the progression. Thus, if the progression contains one perfect \\( k \\) th power, it contains infinitely many.",
  "vars": [
    "n"
  ],
  "params": [
    "a",
    "d",
    "k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "counter",
        "a": "firstterm",
        "d": "commondiff",
        "k": "poweridx"
      },
      "question": "4. Consider the arithmetic progression \\( firstterm, firstterm+commondiff, firstterm+2 commondiff, \\ldots \\), where \\( firstterm \\) and \\( commondiff \\) are positive integers. For any positive integer \\( poweridx \\), prove that the progression has either no exact \\( poweridx \\) th powers or infinitely many.",
      "solution": "Solution. Suppose the given arithmetic progression contains \\( counter^{poweridx} \\) (i.e., it contains a perfect \\( poweridx \\) th power). Then the equation\n\\[\n(counter+commondiff)^{poweridx}=counter^{poweridx}+commondiff\\left|\\binom{poweridx}{1} counter^{poweridx} 1+\\binom{poweridx}{2} counter^{poweridx} 2 commondiff+\\cdots+commondiff^{poweridx} 1\\right|\n\\]\nshows that \\( (counter+commondiff)^{poweridx} \\) is in the progression. By induction, \\( (counter+2 commondiff)^{poweridx} \\), \\( (counter+3 commondiff)^{poweridx}, \\ldots \\) are also in the progression. Thus, if the progression contains one perfect \\( poweridx \\) th power, it contains infinitely many."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "decoration",
        "a": "harboring",
        "d": "locomotion",
        "k": "sandcastle"
      },
      "question": "4. Consider the arithmetic progression \\( harboring, harboring+locomotion, harboring+2 locomotion, \\ldots \\), where \\( harboring \\) and \\( locomotion \\) are positive integers. For any positive integer \\( sandcastle \\), prove that the progression has either no exact \\( sandcastle \\) th powers or infinitely many.",
      "solution": "Solution. Suppose the given arithmetic progression contains \\( decoration^{sandcastle} \\) (i.e., it contains a perfect \\( sandcastle \\) th power). Then the equation\n\\[\n(decoration+locomotion)^{sandcastle}=decoration^{sandcastle}+locomotion\\left|\\binom{sandcastle}{1} decoration^{sandcastle} 1+\\binom{sandcastle}{2} decoration^{sandcastle} 2 locomotion+\\cdots+locomotion^{sandcastle} 1\\right|\n\\]\nshows that \\( (decoration+locomotion)^{sandcastle} \\) is in the progression. By induction, \\( (decoration+2 locomotion)^{sandcastle} \\), \\( (decoration+3 locomotion)^{sandcastle}, \\ldots \\) are also in the progression. Thus, if the progression contains one perfect \\( sandcastle \\) th power, it contains infinitely many."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "finiteval",
        "a": "endvalue",
        "d": "sumvalue",
        "k": "rootvalue"
      },
      "question": "4. Consider the arithmetic progression \\( endvalue, endvalue+sumvalue, endvalue+2 sumvalue, \\ldots \\), where \\( endvalue \\) and \\( sumvalue \\) are positive integers. For any positive integer \\( rootvalue \\), prove that the progression has either no exact \\( rootvalue \\) th powers or infinitely many.",
      "solution": "Solution. Suppose the given arithmetic progression contains \\( finiteval^{rootvalue} \\) (i.e., it contains a perfect \\( rootvalue \\) th power). Then the equation\n\\[\n(finiteval+sumvalue)^{rootvalue}=finiteval^{rootvalue}+sumvalue\\left|\\binom{rootvalue}{1} finiteval^{rootvalue} 1+\\binom{rootvalue}{2} finiteval^{rootvalue} 2 sumvalue+\\cdots+sumvalue^{rootvalue} 1\\right|\n\\]\nshows that \\( (finiteval+sumvalue)^{rootvalue} \\) is in the progression. By induction, \\( (finiteval+2 sumvalue)^{rootvalue} \\), \\( (finiteval+3 sumvalue)^{rootvalue}, \\ldots \\) are also in the progression. Thus, if the progression contains one perfect \\( rootvalue \\) th power, it contains infinitely many."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "a": "pqlmskjt",
        "d": "mxtcrhgz",
        "k": "fvwhqzbo"
      },
      "question": "4. Consider the arithmetic progression \\( pqlmskjt, pqlmskjt+mxtcrhgz, pqlmskjt+2 mxtcrhgz, \\ldots \\), where \\( pqlmskjt \\) and \\( mxtcrhgz \\) are positive integers. For any positive integer \\( fvwhqzbo \\), prove that the progression has either no exact \\( fvwhqzbo \\) th powers or infinitely many.",
      "solution": "Solution. Suppose the given arithmetic progression contains \\( qzxwvtnp^{fvwhqzbo} \\) (i.e., it contains a perfect \\( fvwhqzbo \\) th power). Then the equation\n\\[\n(qzxwvtnp+mxtcrhgz)^{fvwhqzbo}=qzxwvtnp^{fvwhqzbo}+mxtcrhgz\\left|\\binom{fvwhqzbo}{1} qzxwvtnp^{fvwhqzbo} 1+\\binom{fvwhqzbo}{2} qzxwvtnp^{fvwhqzbo} 2 mxtcrhgz+\\cdots+mxtcrhgz^{fvwhqzbo} 1\\right|\n\\]\nshows that \\( (qzxwvtnp+mxtcrhgz)^{fvwhqzbo} \\) is in the progression. By induction, \\( (qzxwvtnp+2 mxtcrhgz)^{fvwhqzbo} \\), \\( (qzxwvtnp+3 mxtcrhgz)^{fvwhqzbo}, \\ldots \\) are also in the progression. Thus, if the progression contains one perfect \\( fvwhqzbo \\) th power, it contains infinitely many."
    },
    "kernel_variant": {
      "question": "Let $K$ be a number field of degree $n\\ge 2$ with ring of integers $\\mathcal{O}_{K}$.  \nFix an element $\\beta\\in\\mathcal{O}_{K}$ that is neither $0$ nor a unit, and an integer $k\\ge 2$.  \nFor $\\alpha\\in\\mathcal{O}_{K}$ consider the $\\mathcal{O}_{K}$–arithmetic progression  \n\\[\n\\mathcal{P}(\\alpha,\\beta):=\\{\\alpha+\\beta\\xi:\\,\\xi\\in\\mathcal{O}_{K}\\}.\n\\]\n\n1.  Prove that the set of $k$-th powers contained in $\\mathcal{P}(\\alpha,\\beta)$ is either empty or infinite; in the latter case one can find infinitely many $k$-th powers whose bases are pair-wise non-associate (that is, no two of the bases differ by multiplication with a unit of $\\mathcal{O}_{K}$).\n\n2.  Assume there exist \\emph{non-zero} elements $\\gamma_{0},\\xi_{0}\\in\\mathcal{O}_{K}$ such that  \n\\[\n\\gamma_{0}^{k}=\\alpha+\\beta\\xi_{0}.\n\\]\nShow that for every non-zero prime ideal $\\mathfrak p\\mid\\beta$ there are infinitely many elements $\\gamma\\in\\mathcal{O}_{K}$ with $\\gamma^{k}\\in\\mathcal{P}(\\alpha,\\beta)$ satisfying  \n\\[\nv_{\\mathfrak p}(\\gamma)=v_{\\mathfrak p}(\\gamma_{0}),\n\\]\nwhere $v_{\\mathfrak p}$ denotes the additive $\\mathfrak p$-adic valuation on $\\mathcal{O}_{K}$, normalised by $v_{\\mathfrak p}(\\mathfrak p)=1$.\n\n(Part 2 refines Part&nbsp;1 by imposing the additional local condition at every prime dividing $\\beta$.)\n\n--------------------------------------------------------------------",
      "solution": "\\textbf{Notation.}  \n$\\bullet$ Two algebraic integers are \\emph{associate} if they differ by multiplication with a unit; we write $\\gamma\\sim\\gamma'$ in this case.  \n$\\bullet$ For a non-zero prime ideal $\\mathfrak p\\subseteq\\mathcal{O}_{K}$ let $v_{\\mathfrak p}$ be the additive valuation with $v_{\\mathfrak p}(\\mathfrak p)=1$.  \n$\\bullet$ Write the prime ideal factorisation  \n\\[\n\\beta=u\\prod_{\\mathfrak p\\mid\\beta}\\mathfrak p^{e_{\\mathfrak p}},\\qquad u\\in\\mathcal{O}_{K}^{\\times},\\;e_{\\mathfrak p}\\ge 1.\n\\]\n\n\\medskip\n\\textbf{A. The progression contains either no $k$-th power or infinitely many.}\n\n\\emph{A1. The divisible case $\\beta\\mid\\alpha$.}  \nWrite $\\alpha=\\beta\\xi_{*}$ with $\\xi_{*}\\in\\mathcal{O}_{K}$.  \nFix a prime ideal $\\mathfrak q$ with $\\mathfrak q\\nmid\\beta$ and let $\\pi\\in\\mathfrak q$ be a uniformiser, i.e.\\ $v_{\\mathfrak q}(\\pi)=1$.  \nFor every integer $m\\ge 1$ put  \n\\[\n\\zeta_{m}:=\\pi^{m},\\qquad \n\\gamma_{m}:=\\beta\\zeta_{m}\\;(=\\beta\\pi^{m}).\n\\]\nThen\n\\[\n\\gamma_{m}^{k}\n      =\\beta^{k}\\pi^{mk}\n      =\\alpha+\\beta\\bigl(\\beta^{\\,k-1}\\pi^{mk}-\\xi_{*}\\bigr)\n      \\in\\mathcal{P}(\\alpha,\\beta)\\quad(m\\ge 1).\n\\]\n\\emph{Non-associativity.}  \nSince $\\mathfrak q\\nmid\\beta$, we have $v_{\\mathfrak q}(\\gamma_{m})=m+v_{\\mathfrak q}(\\beta)=m$.  For $m>m'$,\n\\[\nv_{\\mathfrak q}\\!\\bigl(\\gamma_{m}/\\gamma_{m'}\\bigr)=v_{\\mathfrak q}(\\gamma_{m})-v_{\\mathfrak q}(\\gamma_{m'})=m-m'>0,\n\\]\nhence $\\gamma_{m}/\\gamma_{m'}\\notin\\mathcal{O}_{K}^{\\times}$ and $\\gamma_{m}\\not\\sim\\gamma_{m'}$.  Consequently the divisible case already supplies infinitely many pair-wise non-associate $k$-th powers in $\\mathcal{P}(\\alpha,\\beta)$.\n\n\\emph{A2. The case where a \\emph{non-zero} $k$-th power occurs.}  \nAssume there exist $\\gamma_{0}\\neq 0$ and $\\xi_{0}$ with  \n\\[\n\\gamma_{0}^{k}=\\alpha+\\beta\\xi_{0}. \\tag{1}\n\\]\nFor arbitrary $\\eta\\in\\mathcal{O}_{K}$ define  \n\\[\n\\gamma_{\\eta}:=\\gamma_{0}+\\beta\\eta. \\tag{2}\n\\]\nBy the binomial theorem (valid in any Dedekind domain)\n\\[\n\\gamma_{\\eta}^{k}\n      =\\gamma_{0}^{k}+\\beta\\Phi(\\eta), \\tag{3}\n\\]\nwhere $\\Phi\\in\\mathcal{O}_{K}[X]$ has $\\Phi(0)=0$.  Combining (1) and (3) gives\n\\[\n\\gamma_{\\eta}^{k}=\\alpha+\\beta\\bigl(\\xi_{0}+\\Phi(\\eta)\\bigr)\\in\\mathcal{P}(\\alpha,\\beta). \\tag{4}\n\\]\nThus $\\mathcal{P}(\\alpha,\\beta)$ contains infinitely many $k$-th powers indexed by $\\eta$.\n\nSo far we have established that \\emph{either} no $k$-th power occurs \\emph{or} there are infinitely many.  To finish Part 1 we still need pair-wise non-associate bases in the non-divisible situation; Parts B and C provide them while simultaneously proving Part 2.\n\n\\medskip\n\\textbf{B. Controlling the valuations at the primes dividing $\\beta$.}\n\nFix the witnessing element $\\gamma_{0}$ from (1).  For every $\\mathfrak p\\mid\\beta$ set  \n\\[\nM_{\\mathfrak p}:=\\max\\bigl\\{1,\\;v_{\\mathfrak p}(\\gamma_{0})-e_{\\mathfrak p}+1\\bigr\\},\\qquad\n\\mathfrak b:=\\prod_{\\mathfrak p\\mid\\beta}\\mathfrak p^{M_{\\mathfrak p}}. \\tag{5}\n\\]\nIf $\\eta\\in\\mathfrak b$ then  \n\\[\nv_{\\mathfrak p}(\\beta\\eta)\n      =e_{\\mathfrak p}+v_{\\mathfrak p}(\\eta)\n      >e_{\\mathfrak p}+v_{\\mathfrak p}(\\gamma_{0})-e_{\\mathfrak p}=v_{\\mathfrak p}(\\gamma_{0})\n      \\quad(\\mathfrak p\\mid\\beta), \\tag{6}\n\\]\nhence  \n\\[\nv_{\\mathfrak p}(\\gamma_{0}+\\beta\\eta)=v_{\\mathfrak p}(\\gamma_{0})\\quad(\\mathfrak p\\mid\\beta,\\;\\eta\\in\\mathfrak b). \\tag{7}\n\\]\n\n\\medskip\n\\textbf{C. Construction of pair-wise non-associate bases.}\n\n\\emph{C1. A convenient family.}  \nChoose any $\\eta_{*}\\in\\mathfrak b\\setminus\\{0\\}$ and define, for $m=1,2,3,\\dots$,\n\\[\n\\eta_{m}:=m\\eta_{*},\\qquad\n\\gamma_{m}:=\\gamma_{0}+\\beta\\eta_{m}. \\tag{8}\n\\]\nBy (7) we get  \n\\[\nv_{\\mathfrak p}(\\gamma_{m})=v_{\\mathfrak p}(\\gamma_{0})\n\\quad\\Bigl(\\mathfrak p\\mid\\beta,\\;m\\ge 1\\Bigr). \\tag{9}\n\\]\nThus every $\\gamma_{m}$ already fulfils the local condition required in Part 2.\n\n\\emph{C2. Strict growth of the absolute norm.}  \nConsider the norm polynomial\n\\[\nF(X):=N_{K/\\mathbb{Q}}\\bigl(\\gamma_{0}+\\beta\\eta_{*}X\\bigr)\\in\\mathbb{Z}[X]. \\tag{10}\n\\]\n(The coefficients are elementary symmetric polynomials in the algebraic integers $\\sigma(\\gamma_{0}+\\beta\\eta_{*}X)$, hence belong to $\\mathbb{Z}$.)  \n$F$ has degree $n$ and leading coefficient \n\\[\nc:=N_{K/\\mathbb{Q}}(\\beta\\eta_{*})\\neq 0,\n\\]\nso $\\lvert F(t)\\rvert\\to\\infty$ as $\\lvert t\\rvert\\to\\infty$.  The finite set  \n\\[\n\\mathcal{Z}:=\\{t\\in\\mathbb{Z}\\mid F(t)=0\\}\n\\]\ndoes not contain all positive integers.  Pick an infinite, strictly increasing sequence of positive integers\n\\[\n1\\le m_{1}<m_{2}<m_{3}<\\dots,\\qquad m_{r}\\not\\in\\mathcal{Z},\\qquad\n\\lvert F(m_{1})\\rvert<\\lvert F(m_{2})\\rvert<\\lvert F(m_{3})\\rvert<\\dots \\tag{11}\n\\]\nand put $\\gamma_{(r)}:=\\gamma_{m_{r}}$.\n\n\\emph{C3. Excluding associativity.}  \nFirst note that $N_{K/\\mathbb{Q}}\\bigl(\\gamma_{(r)}\\bigr)=F(m_{r})\\neq 0$, so $\\gamma_{(r)}\\neq 0$.  If $\\gamma_{(r)}\\sim\\gamma_{(s)}$ with $r<s$, then $\\gamma_{(r)}=\\varepsilon\\gamma_{(s)}$ for some unit $\\varepsilon$, and therefore\n\\[\n\\lvert F(m_{r})\\rvert\n      =\\lvert N_{K/\\mathbb{Q}}(\\varepsilon)\\rvert\\,\\lvert F(m_{s})\\rvert\n      =\\lvert F(m_{s})\\rvert,\n\\]\nbecause $N_{K/\\mathbb{Q}}(\\varepsilon)=\\pm1$.  This contradicts (11); hence $\\gamma_{(r)}\\not\\sim\\gamma_{(s)}$ for $r\\neq s$.\n\n\\emph{C4. Membership in the progression.}  \nFor each $r$,\n\\[\n\\gamma_{(r)}^{k}\n     =\\gamma_{m_{r}}^{k}\n     =\\gamma_{0}^{k}+\\beta\\Phi(\\eta_{m_{r}})     \\quad\\text{by}\\;(3)\n     =\\alpha+\\beta\\bigl(\\xi_{0}+\\Phi(\\eta_{m_{r}})\\bigr)\n     \\in\\mathcal{P}(\\alpha,\\beta),\n\\]\nso $\\{\\gamma_{(r)}^{k}\\mid r\\ge 1\\}$ is an infinite subset of $k$-th powers in $\\mathcal{P}(\\alpha,\\beta)$ whose bases are pair-wise non-associate and satisfy (9), i.e.\\ condition $(\\dagger)$, at every $\\mathfrak p\\mid\\beta$.\n\n\\medskip\n\\textbf{D. Conclusion.}\n\n\\begin{itemize}\n\\item If $\\mathcal{P}(\\alpha,\\beta)$ contains no $k$-th power, Part 1 is vacuously true.\n\\item If $\\beta\\mid\\alpha$, Section A1 supplies the required infinite family for both parts of the problem.\n\\item Otherwise, Sections A2–C construct an infinite set of $k$-th powers in $\\mathcal{P}(\\alpha,\\beta)$ whose bases are pair-wise non-associate and satisfy the valuation constraint at all $\\mathfrak p\\mid\\beta$, thereby proving Parts 1 and 2 simultaneously.\n\\end{itemize}\n\\qed\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.523891",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-dimensional setting: the problem moves from ℤ to an arbitrary number field K with degree n ≥ 2, so one must work inside the Dedekind domain 𝒪ₖ rather than the PID ℤ.  \n• Additional structure: prime-ideal factorizations, 𝔭-adic valuations, units and associates must all be handled explicitly.  \n• Extra constraints: Part 2 requires keeping each v_𝔭(γ) fixed while still creating infinitely many new solutions, which introduces non-trivial local-global (𝔭-adic vs. Archimedean) considerations.  \n• Deeper theory: the solution employs Dedekind-domain properties, non-Archimedean valuations, and unit groups, far beyond the elementary binomial-expansion argument sufficient for the original exercise.  \n• Length and sophistication: each step (existence → infinitude, control of associates, prime-wise valuation management, norm growth) adds layers of reasoning, making the enhanced variant markedly harder than both the original problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $K$ be a number field of degree $n\\ge 2$ with ring of integers $\\mathcal{O}_{K}$.  \nFix an element $\\beta\\in\\mathcal{O}_{K}$ that is neither $0$ nor a unit, and an integer $k\\ge 2$.  \nFor $\\alpha\\in\\mathcal{O}_{K}$ consider the $\\mathcal{O}_{K}$–arithmetic progression  \n\\[\n\\mathcal{P}(\\alpha,\\beta):=\\{\\alpha+\\beta\\xi:\\,\\xi\\in\\mathcal{O}_{K}\\}.\n\\]\n\n1.  Prove that the set of $k$-th powers contained in $\\mathcal{P}(\\alpha,\\beta)$ is either empty or infinite; in the latter case one can find infinitely many $k$-th powers whose bases are pair-wise non-associate (that is, no two of the bases differ by multiplication with a unit of $\\mathcal{O}_{K}$).\n\n2.  Assume there exist \\emph{non-zero} elements $\\gamma_{0},\\xi_{0}\\in\\mathcal{O}_{K}$ such that  \n\\[\n\\gamma_{0}^{k}=\\alpha+\\beta\\xi_{0}.\n\\]\nShow that for every non-zero prime ideal $\\mathfrak p\\mid\\beta$ there are infinitely many elements $\\gamma\\in\\mathcal{O}_{K}$ with $\\gamma^{k}\\in\\mathcal{P}(\\alpha,\\beta)$ satisfying  \n\\[\nv_{\\mathfrak p}(\\gamma)=v_{\\mathfrak p}(\\gamma_{0}),\n\\]\nwhere $v_{\\mathfrak p}$ denotes the additive $\\mathfrak p$-adic valuation on $\\mathcal{O}_{K}$, normalised by $v_{\\mathfrak p}(\\mathfrak p)=1$.\n\n(Part 2 refines Part&nbsp;1 by imposing the additional local condition at every prime dividing $\\beta$.)\n\n--------------------------------------------------------------------",
      "solution": "\\textbf{Notation.}  \n$\\bullet$ Two algebraic integers are \\emph{associate} if they differ by multiplication with a unit; we write $\\gamma\\sim\\gamma'$ in this case.  \n$\\bullet$ For a non-zero prime ideal $\\mathfrak p\\subseteq\\mathcal{O}_{K}$ let $v_{\\mathfrak p}$ be the additive valuation with $v_{\\mathfrak p}(\\mathfrak p)=1$.  \n$\\bullet$ Write the prime ideal factorisation  \n\\[\n\\beta=u\\prod_{\\mathfrak p\\mid\\beta}\\mathfrak p^{e_{\\mathfrak p}},\\qquad u\\in\\mathcal{O}_{K}^{\\times},\\;e_{\\mathfrak p}\\ge 1.\n\\]\n\n\\medskip\n\\textbf{A. The progression contains either no $k$-th power or infinitely many.}\n\n\\emph{A1. The divisible case $\\beta\\mid\\alpha$.}  \nWrite $\\alpha=\\beta\\xi_{*}$ with $\\xi_{*}\\in\\mathcal{O}_{K}$.  \nFix a prime ideal $\\mathfrak q$ with $\\mathfrak q\\nmid\\beta$ and let $\\pi\\in\\mathfrak q$ be a uniformiser, i.e.\\ $v_{\\mathfrak q}(\\pi)=1$.  \nFor every integer $m\\ge 1$ put  \n\\[\n\\zeta_{m}:=\\pi^{m},\\qquad \n\\gamma_{m}:=\\beta\\zeta_{m}\\;(=\\beta\\pi^{m}).\n\\]\nThen\n\\[\n\\gamma_{m}^{k}\n      =\\beta^{k}\\pi^{mk}\n      =\\alpha+\\beta\\bigl(\\beta^{\\,k-1}\\pi^{mk}-\\xi_{*}\\bigr)\n      \\in\\mathcal{P}(\\alpha,\\beta)\\quad(m\\ge 1).\n\\]\n\\emph{Non-associativity.}  \nSince $\\mathfrak q\\nmid\\beta$, we have $v_{\\mathfrak q}(\\gamma_{m})=m+v_{\\mathfrak q}(\\beta)=m$.  For $m>m'$,\n\\[\nv_{\\mathfrak q}\\!\\bigl(\\gamma_{m}/\\gamma_{m'}\\bigr)=v_{\\mathfrak q}(\\gamma_{m})-v_{\\mathfrak q}(\\gamma_{m'})=m-m'>0,\n\\]\nhence $\\gamma_{m}/\\gamma_{m'}\\notin\\mathcal{O}_{K}^{\\times}$ and $\\gamma_{m}\\not\\sim\\gamma_{m'}$.  Consequently the divisible case already supplies infinitely many pair-wise non-associate $k$-th powers in $\\mathcal{P}(\\alpha,\\beta)$.\n\n\\emph{A2. The case where a \\emph{non-zero} $k$-th power occurs.}  \nAssume there exist $\\gamma_{0}\\neq 0$ and $\\xi_{0}$ with  \n\\[\n\\gamma_{0}^{k}=\\alpha+\\beta\\xi_{0}. \\tag{1}\n\\]\nFor arbitrary $\\eta\\in\\mathcal{O}_{K}$ define  \n\\[\n\\gamma_{\\eta}:=\\gamma_{0}+\\beta\\eta. \\tag{2}\n\\]\nBy the binomial theorem (valid in any Dedekind domain)\n\\[\n\\gamma_{\\eta}^{k}\n      =\\gamma_{0}^{k}+\\beta\\Phi(\\eta), \\tag{3}\n\\]\nwhere $\\Phi\\in\\mathcal{O}_{K}[X]$ has $\\Phi(0)=0$.  Combining (1) and (3) gives\n\\[\n\\gamma_{\\eta}^{k}=\\alpha+\\beta\\bigl(\\xi_{0}+\\Phi(\\eta)\\bigr)\\in\\mathcal{P}(\\alpha,\\beta). \\tag{4}\n\\]\nThus $\\mathcal{P}(\\alpha,\\beta)$ contains infinitely many $k$-th powers indexed by $\\eta$.\n\nSo far we have established that \\emph{either} no $k$-th power occurs \\emph{or} there are infinitely many.  To finish Part 1 we still need pair-wise non-associate bases in the non-divisible situation; Parts B and C provide them while simultaneously proving Part 2.\n\n\\medskip\n\\textbf{B. Controlling the valuations at the primes dividing $\\beta$.}\n\nFix the witnessing element $\\gamma_{0}$ from (1).  For every $\\mathfrak p\\mid\\beta$ set  \n\\[\nM_{\\mathfrak p}:=\\max\\bigl\\{1,\\;v_{\\mathfrak p}(\\gamma_{0})-e_{\\mathfrak p}+1\\bigr\\},\\qquad\n\\mathfrak b:=\\prod_{\\mathfrak p\\mid\\beta}\\mathfrak p^{M_{\\mathfrak p}}. \\tag{5}\n\\]\nIf $\\eta\\in\\mathfrak b$ then  \n\\[\nv_{\\mathfrak p}(\\beta\\eta)\n      =e_{\\mathfrak p}+v_{\\mathfrak p}(\\eta)\n      >e_{\\mathfrak p}+v_{\\mathfrak p}(\\gamma_{0})-e_{\\mathfrak p}=v_{\\mathfrak p}(\\gamma_{0})\n      \\quad(\\mathfrak p\\mid\\beta), \\tag{6}\n\\]\nhence  \n\\[\nv_{\\mathfrak p}(\\gamma_{0}+\\beta\\eta)=v_{\\mathfrak p}(\\gamma_{0})\\quad(\\mathfrak p\\mid\\beta,\\;\\eta\\in\\mathfrak b). \\tag{7}\n\\]\n\n\\medskip\n\\textbf{C. Construction of pair-wise non-associate bases.}\n\n\\emph{C1. A convenient family.}  \nChoose any $\\eta_{*}\\in\\mathfrak b\\setminus\\{0\\}$ and define, for $m=1,2,3,\\dots$,\n\\[\n\\eta_{m}:=m\\eta_{*},\\qquad\n\\gamma_{m}:=\\gamma_{0}+\\beta\\eta_{m}. \\tag{8}\n\\]\nBy (7) we get  \n\\[\nv_{\\mathfrak p}(\\gamma_{m})=v_{\\mathfrak p}(\\gamma_{0})\n\\quad\\Bigl(\\mathfrak p\\mid\\beta,\\;m\\ge 1\\Bigr). \\tag{9}\n\\]\nThus every $\\gamma_{m}$ already fulfils the local condition required in Part 2.\n\n\\emph{C2. Strict growth of the absolute norm.}  \nConsider the norm polynomial\n\\[\nF(X):=N_{K/\\mathbb{Q}}\\bigl(\\gamma_{0}+\\beta\\eta_{*}X\\bigr)\\in\\mathbb{Z}[X]. \\tag{10}\n\\]\n(The coefficients are elementary symmetric polynomials in the algebraic integers $\\sigma(\\gamma_{0}+\\beta\\eta_{*}X)$, hence belong to $\\mathbb{Z}$.)  \n$F$ has degree $n$ and leading coefficient \n\\[\nc:=N_{K/\\mathbb{Q}}(\\beta\\eta_{*})\\neq 0,\n\\]\nso $\\lvert F(t)\\rvert\\to\\infty$ as $\\lvert t\\rvert\\to\\infty$.  The finite set  \n\\[\n\\mathcal{Z}:=\\{t\\in\\mathbb{Z}\\mid F(t)=0\\}\n\\]\ndoes not contain all positive integers.  Pick an infinite, strictly increasing sequence of positive integers\n\\[\n1\\le m_{1}<m_{2}<m_{3}<\\dots,\\qquad m_{r}\\not\\in\\mathcal{Z},\\qquad\n\\lvert F(m_{1})\\rvert<\\lvert F(m_{2})\\rvert<\\lvert F(m_{3})\\rvert<\\dots \\tag{11}\n\\]\nand put $\\gamma_{(r)}:=\\gamma_{m_{r}}$.\n\n\\emph{C3. Excluding associativity.}  \nFirst note that $N_{K/\\mathbb{Q}}\\bigl(\\gamma_{(r)}\\bigr)=F(m_{r})\\neq 0$, so $\\gamma_{(r)}\\neq 0$.  If $\\gamma_{(r)}\\sim\\gamma_{(s)}$ with $r<s$, then $\\gamma_{(r)}=\\varepsilon\\gamma_{(s)}$ for some unit $\\varepsilon$, and therefore\n\\[\n\\lvert F(m_{r})\\rvert\n      =\\lvert N_{K/\\mathbb{Q}}(\\varepsilon)\\rvert\\,\\lvert F(m_{s})\\rvert\n      =\\lvert F(m_{s})\\rvert,\n\\]\nbecause $N_{K/\\mathbb{Q}}(\\varepsilon)=\\pm1$.  This contradicts (11); hence $\\gamma_{(r)}\\not\\sim\\gamma_{(s)}$ for $r\\neq s$.\n\n\\emph{C4. Membership in the progression.}  \nFor each $r$,\n\\[\n\\gamma_{(r)}^{k}\n     =\\gamma_{m_{r}}^{k}\n     =\\gamma_{0}^{k}+\\beta\\Phi(\\eta_{m_{r}})     \\quad\\text{by}\\;(3)\n     =\\alpha+\\beta\\bigl(\\xi_{0}+\\Phi(\\eta_{m_{r}})\\bigr)\n     \\in\\mathcal{P}(\\alpha,\\beta),\n\\]\nso $\\{\\gamma_{(r)}^{k}\\mid r\\ge 1\\}$ is an infinite subset of $k$-th powers in $\\mathcal{P}(\\alpha,\\beta)$ whose bases are pair-wise non-associate and satisfy (9), i.e.\\ condition $(\\dagger)$, at every $\\mathfrak p\\mid\\beta$.\n\n\\medskip\n\\textbf{D. Conclusion.}\n\n\\begin{itemize}\n\\item If $\\mathcal{P}(\\alpha,\\beta)$ contains no $k$-th power, Part 1 is vacuously true.\n\\item If $\\beta\\mid\\alpha$, Section A1 supplies the required infinite family for both parts of the problem.\n\\item Otherwise, Sections A2–C construct an infinite set of $k$-th powers in $\\mathcal{P}(\\alpha,\\beta)$ whose bases are pair-wise non-associate and satisfy the valuation constraint at all $\\mathfrak p\\mid\\beta$, thereby proving Parts 1 and 2 simultaneously.\n\\end{itemize}\n\\qed\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.438195",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-dimensional setting: the problem moves from ℤ to an arbitrary number field K with degree n ≥ 2, so one must work inside the Dedekind domain 𝒪ₖ rather than the PID ℤ.  \n• Additional structure: prime-ideal factorizations, 𝔭-adic valuations, units and associates must all be handled explicitly.  \n• Extra constraints: Part 2 requires keeping each v_𝔭(γ) fixed while still creating infinitely many new solutions, which introduces non-trivial local-global (𝔭-adic vs. Archimedean) considerations.  \n• Deeper theory: the solution employs Dedekind-domain properties, non-Archimedean valuations, and unit groups, far beyond the elementary binomial-expansion argument sufficient for the original exercise.  \n• Length and sophistication: each step (existence → infinitude, control of associates, prime-wise valuation management, norm growth) adds layers of reasoning, making the enhanced variant markedly harder than both the original problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}