{
  "index": "2013-A-2",
  "type": "NT",
  "tag": [
    "NT",
    "COMB"
  ],
  "difficulty": "",
  "question": "Let $S$ be the set of all positive integers that are \\emph{not} perfect squares. For $n$ in $S$, consider choices of integers\n$a_1, a_2, \\dots, a_r$ such that $n < a_1<  a_2 < \\cdots < a_r$\nand $n \\cdot a_1 \\cdot a_2 \\cdots a_r$ is a perfect square, and\nlet $f(n)$ be the minumum of $a_r$ over all such choices. For example,\n$2 \\cdot 3 \\cdot 6$ is a perfect square, while $2 \\cdot 3$, $2 \\cdot 4$, \n$2 \\cdot 5$, $2 \\cdot 3 \\cdot 4$, $2 \\cdot 3 \\cdot 5$, $2 \\cdot 4 \\cdot 5$, and $2 \\cdot 3 \\cdot 4 \\cdot 5$ are not, and so $f(2) = 6$.\nShow that the function $f$ from $S$ to the integers is one-to-one.",
  "solution": "Suppose to the contrary that $f(n) = f(m)$ with $n<m$, and let $n\\cdot a_1\\cdots a_r$, $m\\cdot b_1\\cdots b_s$ be perfect squares where $n < a_1 < \\cdots < a_r$, $m < b_1 < \\cdots < b_s$, $a_r,b_s$ are minimal and $a_r=b_s$. Then $(n\\cdot a_1\\cdots a_r)\\cdot (m\\cdot b_1\\cdots b_s)$ is also a perfect square. Now eliminate any factor in this product that appears twice (i.e., if $a_i = b_j$ for some $i,j$, then delete $a_i$ and $b_j$ from this product). The product of what remains must also be a perfect square, but this is now a product of distinct integers, the smallest of which is $n$ and the largest of which is strictly smaller than $a_r = b_s$. This contradicts the minimality of $a_r$.\n\n\\noindent\n\\textbf{Remark:}\nSequences whose product is a perfect square occur naturally in the \\emph{quadratic sieve} algorithm for factoring large integers. However,\nthe behavior of the function $f(n)$ seems to be somewhat erratic. \nKarl Mahlburg points out the upper bound $f(n) \\leq 2n$ for $n \\geq 5$, which holds because the interval $(n, 2n)$ contains an integer of the form $2m^2$. A trivial lower bound is $f(n) \\geq n+p$ where $p$ is the least prime factor of $n$. For $n = p$ prime, the bounds agree and we have $f(p) = 2p$. For more discussion, see \n\\url{https://oeis.org/A006255}.",
  "vars": [
    "S",
    "n",
    "a_1",
    "a_2",
    "a_r",
    "r",
    "m",
    "b_1",
    "b_s",
    "s",
    "i",
    "j",
    "f",
    "p"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "S": "nonsquareset",
        "n": "baseint",
        "a_1": "firstauxint",
        "a_2": "secondauxint",
        "a_r": "lastauxint",
        "r": "auxcount",
        "m": "compint",
        "b_1": "firstbeta",
        "b_s": "lastbeta",
        "s": "betacount",
        "j": "secondidx",
        "f": "minimaxfunc",
        "p": "leastprime"
      },
      "question": "Let $nonsquareset$ be the set of all positive integers that are \\emph{not} perfect squares. For $baseint$ in $nonsquareset$, consider choices of integers\n$firstauxint, secondauxint, \\dots, lastauxint$ such that $baseint < firstauxint<  secondauxint < \\cdots < lastauxint$\nand $baseint \\cdot firstauxint \\cdot secondauxint \\cdots lastauxint$ is a perfect square, and\nlet $minimaxfunc(baseint)$ be the minumum of $lastauxint$ over all such choices. For example,\n$2 \\cdot 3 \\cdot 6$ is a perfect square, while $2 \\cdot 3$, $2 \\cdot 4$, \n$2 \\cdot 5$, $2 \\cdot 3 \\cdot 4$, $2 \\cdot 3 \\cdot 5$, $2 \\cdot 4 \\cdot 5$, and $2 \\cdot 3 \\cdot 4 \\cdot 5$ are not, and so $minimaxfunc(2) = 6$.\nShow that the function $minimaxfunc$ from $nonsquareset$ to the integers is one-to-one.",
      "solution": "Suppose to the contrary that $minimaxfunc(baseint) = minimaxfunc(compint)$ with $baseint<compint$, and let $baseint\\cdot firstauxint\\cdots lastauxint$, $compint\\cdot firstbeta\\cdots lastbeta$ be perfect squares where $baseint < firstauxint < \\cdots < lastauxint$, $compint < firstbeta < \\cdots < lastbeta$, $lastauxint,lastbeta$ are minimal and $lastauxint=lastbeta$. Then $(baseint\\cdot firstauxint\\cdots lastauxint)\\cdot (compint\\cdot firstbeta\\cdots lastbeta)$ is also a perfect square. Now eliminate any factor in this product that appears twice (i.e., if $a_i = b_{secondidx}$ for some $i,secondidx$, then delete $a_i$ and $b_{secondidx}$ from this product). The product of what remains must also be a perfect square, but this is now a product of distinct integers, the smallest of which is $baseint$ and the largest of which is strictly smaller than $lastauxint = lastbeta$. This contradicts the minimality of $lastauxint$.\n\n\\noindent\n\\textbf{Remark:}\nSequences whose product is a perfect square occur naturally in the \\emph{quadratic sieve} algorithm for factoring large integers. However,\nthe behavior of the function $minimaxfunc(baseint)$ seems to be somewhat erratic. \nKarl Mahlburg points out the upper bound $minimaxfunc(baseint) \\leq 2baseint$ for $baseint \\geq 5$, which holds because the interval $(baseint, 2baseint)$ contains an integer of the form $2compint^2$. A trivial lower bound is $minimaxfunc(baseint) \\geq baseint+leastprime$ where $leastprime$ is the least prime factor of $baseint$. For $baseint = leastprime$ prime, the bounds agree and we have $minimaxfunc(leastprime) = 2leastprime$. For more discussion, see \n\\url{https://oeis.org/A006255}."
    },
    "descriptive_long_confusing": {
      "map": {
        "S": "riverbank",
        "n": "lighthouse",
        "a_1": "snowflake",
        "a_2": "paintbrush",
        "a_r": "sandcastle",
        "r": "trombone",
        "m": "stargazer",
        "b_1": "moonstone",
        "b_s": "thunderbolt",
        "s": "windchime",
        "i": "harmonica",
        "j": "daydream",
        "f": "waterfall",
        "p": "gemstone"
      },
      "question": "Let $riverbank$ be the set of all positive integers that are \\emph{not} perfect squares. For $lighthouse$ in $riverbank$, consider choices of integers\n$snowflake, paintbrush, \\dots, sandcastle$ such that $lighthouse < snowflake<  paintbrush < \\cdots < sandcastle$\nand $lighthouse \\cdot snowflake \\cdot paintbrush \\cdots sandcastle$ is a perfect square, and\nlet $waterfall(lighthouse)$ be the minumum of $sandcastle$ over all such choices. For example,\n$2 \\cdot 3 \\cdot 6$ is a perfect square, while $2 \\cdot 3$, $2 \\cdot 4$, \n$2 \\cdot 5$, $2 \\cdot 3 \\cdot 4$, $2 \\cdot 3 \\cdot 5$, $2 \\cdot 4 \\cdot 5$, and $2 \\cdot 3 \\cdot 4 \\cdot 5$ are not, and so $waterfall(2) = 6$.\nShow that the function $waterfall$ from $riverbank$ to the integers is one-to-one.",
      "solution": "Suppose to the contrary that $waterfall(lighthouse) = waterfall(stargazer)$ with $lighthouse<stargazer$, and let $lighthouse\\cdot snowflake\\cdots sandcastle$, $stargazer\\cdot moonstone\\cdots thunderbolt$ be perfect squares where $lighthouse < snowflake < \\cdots < sandcastle$, $stargazer < moonstone < \\cdots < thunderbolt$, $sandcastle,thunderbolt$ are minimal and $sandcastle=thunderbolt$. Then $(lighthouse\\cdot snowflake\\cdots sandcastle)\\cdot (stargazer\\cdot moonstone\\cdots thunderbolt)$ is also a perfect square. Now eliminate any factor in this product that appears twice (i.e., if $a_{harmonica} = b_{daydream}$ for some $\\harmonica,\\daydream$, then delete $a_{harmonica}$ and $b_{daydream}$ from this product). The product of what remains must also be a perfect square, but this is now a product of distinct integers, the smallest of which is $lighthouse$ and the largest of which is strictly smaller than $sandcastle = thunderbolt$. This contradicts the minimality of $sandcastle$.",
      "confidence": 0.13
    },
    "descriptive_long_misleading": {
      "map": {
        "S": "emptymultitude",
        "n": "irrationalvalue",
        "a_1": "finalfactor",
        "a_2": "penultimatefactor",
        "a_r": "minimalfactor",
        "r": "infiniteindex",
        "m": "constantvalue",
        "b_1": "finalelement",
        "b_s": "initialfactor",
        "s": "eternalindex",
        "i": "outerindex",
        "j": "staticindex",
        "f": "constantmap",
        "p": "compositefactor"
      },
      "question": "Let $emptymultitude$ be the set of all positive integers that are \\emph{not} perfect squares. For $irrationalvalue$ in $emptymultitude$, consider choices of integers\n$finalfactor, penultimatefactor, \\dots, minimalfactor$ such that $irrationalvalue < finalfactor<  penultimatefactor < \\cdots < minimalfactor$\nand $irrationalvalue \\cdot finalfactor \\cdot penultimatefactor \\cdots minimalfactor$ is a perfect square, and\nlet $constantmap(irrationalvalue)$ be the minumum of $minimalfactor$ over all such choices. For example,\n$2 \\cdot 3 \\cdot 6$ is a perfect square, while $2 \\cdot 3$, $2 \\cdot 4$, \n$2 \\cdot 5$, $2 \\cdot 3 \\cdot 4$, $2 \\cdot 3 \\cdot 5$, $2 \\cdot 4 \\cdot 5$, and $2 \\cdot 3 \\cdot 4 \\cdot 5$ are not, and so $constantmap(2) = 6$.\nShow that the function $constantmap$ from $emptymultitude$ to the integers is one-to-one.",
      "solution": "Suppose to the contrary that $constantmap(irrationalvalue) = constantmap(constantvalue)$ with $irrationalvalue<constantvalue$, and let $irrationalvalue\\cdot finalfactor\\cdots minimalfactor$, $constantvalue\\cdot finalelement\\cdots initialfactor$ be perfect squares where $irrationalvalue < finalfactor < \\cdots < minimalfactor$, $constantvalue < finalelement < \\cdots < initialfactor$, $minimalfactor,initialfactor$ are minimal and $minimalfactor=initialfactor$. Then $(irrationalvalue\\cdot finalfactor\\cdots minimalfactor)\\cdot (constantvalue\\cdot finalelement\\cdots initialfactor)$ is also a perfect square. Now eliminate any factor in this product that appears twice (i.e., if $a_{outerindex} = b_{staticindex}$ for some $outerindex,staticindex$, then delete $a_{outerindex}$ and $b_{staticindex}$ from this product). The product of what remains must also be a perfect square, but this is now a product of distinct integers, the smallest of which is $irrationalvalue$ and the largest of which is strictly smaller than $minimalfactor = initialfactor$. This contradicts the minimality of $minimalfactor$.\n\n\\noindent\n\\textbf{Remark:}\nKarl Mahlburg points out the upper bound $constantmap(irrationalvalue) \\leq 2irrationalvalue$ for $irrationalvalue \\geq 5$, which holds because the interval $(irrationalvalue, 2irrationalvalue)$ contains an integer of the form $2constantvalue^2$. A trivial lower bound is $constantmap(irrationalvalue) \\geq irrationalvalue+compositefactor$ where $compositefactor$ is the least prime factor of $irrationalvalue$. For $irrationalvalue = compositefactor$ prime, the bounds agree and we have $constantmap(compositefactor) = 2compositefactor$. For more discussion, see \n\\url{https://oeis.org/A006255}."
    },
    "garbled_string": {
      "map": {
        "S": "qzxwvtnp",
        "n": "hjgrksla",
        "a_1": "mlfdqxze",
        "a_2": "sjvoknhr",
        "a_r": "ptclzmea",
        "r": "xdukfrno",
        "m": "ginuyhte",
        "b_1": "kpshwrqd",
        "b_s": "uctemxla",
        "s": "zwyrmkef",
        "i": "obklvnaq",
        "j": "wumrcpse",
        "f": "yqdnseite",
        "p": "avhlkjwe"
      },
      "question": "Let $qzxwvtnp$ be the set of all positive integers that are \\emph{not} perfect squares. For $hjgrksla$ in $qzxwvtnp$, consider choices of integers\n$mlfdqxze, sjvoknhr, \\dots, ptclzmea$ such that $hjgrksla < mlfdqxze<  sjvoknhr < \\cdots < ptclzmea$\nand $hjgrksla \\cdot mlfdqxze \\cdot sjvoknhr \\cdots ptclzmea$ is a perfect square, and\nlet $yqdnseite(hjgrksla)$ be the minumum of $ptclzmea$ over all such choices. For example,\n$2 \\cdot 3 \\cdot 6$ is a perfect square, while $2 \\cdot 3$, $2 \\cdot 4$, \n$2 \\cdot 5$, $2 \\cdot 3 \\cdot 4$, $2 \\cdot 3 \\cdot 5$, $2 \\cdot 4 \\cdot 5$, and $2 \\cdot 3 \\cdot 4 \\cdot 5$ are not, and so $yqdnseite(2) = 6$.\nShow that the function $yqdnseite$ from $qzxwvtnp$ to the integers is one-to-one.",
      "solution": "Suppose to the contrary that $yqdnseite(hjgrksla) = yqdnseite(ginuyhte)$ with $hjgrksla<ginuyhte$, and let $hjgrksla\\cdot mlfdqxze\\cdots ptclzmea$, $ginuyhte\\cdot kpshwrqd\\cdots uctemxla$ be perfect squares where $hjgrksla < mlfdqxze < \\cdots < ptclzmea$, $ginuyhte < kpshwrqd < \\cdots < uctemxla$, $ptclzmea,uctemxla$ are minimal and $ptclzmea=uctemxla$. Then $(hjgrksla\\cdot mlfdqxze\\cdots ptclzmea)\\cdot (ginuyhte\\cdot kpshwrqd\\cdots uctemxla)$ is also a perfect square. Now eliminate any factor in this product that appears twice (i.e., if $a_{obklvnaq} = b_{wumrcpse}$ for some $obklvnaq,wumrcpse$, then delete $a_{obklvnaq}$ and $b_{wumrcpse}$ from this product). The product of what remains must also be a perfect square, but this is now a product of distinct integers, the smallest of which is $hjgrksla$ and the largest of which is strictly smaller than $ptclzmea = uctemxla$. This contradicts the minimality of $ptclzmea$. \n\n\\noindent\n\\textbf{Remark:}\nSequences whose product is a perfect square occur naturally in the \\emph{quadratic sieve} algorithm for factoring large integers. However,\nthe behavior of the function $yqdnseite(hjgrksla)$ seems to be somewhat erratic. \nKarl Mahlburg points out the upper bound $yqdnseite(hjgrksla) \\leq 2hjgrksla$ for $hjgrksla \\geq 5$, which holds because the interval $(hjgrksla, 2hjgrksla)$ contains an integer of the form $2m^2$. A trivial lower bound is $yqdnseite(hjgrksla) \\geq hjgrksla+avhlkjwe$ where $avhlkjwe$ is the least prime factor of $hjgrksla$. For $hjgrksla = avhlkjwe$ prime, the bounds agree and we have $yqdnseite(avhlkjwe) = 2avhlkjwe$. For more discussion, see \n\\url{https://oeis.org/A006255}.}"
    },
    "kernel_variant": {
      "question": "Let $k\\ge 2$ be a fixed prime number and put \n\\[\nS_k\\;=\\;\\bigl\\{\\,n\\in\\mathbb N\\;:\\; n\\text{ is \\emph{not} a perfect }k\\text{-th power}\\bigr\\}.\n\\]\n\nFor $n\\in S_k$ call an ordered finite tuple \n\\[\nA=(a_1,\\dots ,a_r)\\qquad(r\\ge 1)\n\\]\n$k$–admissible for $n$ if  \n\n1.\\,(Order)\\quad $n<a_1<a_2<\\dots <a_r$ and the $a_i$ are pair-wise distinct;\n\n2.\\,(Perfection)\\quad the product $n\\cdot a_1a_2\\dots a_r$ is a perfect $k$-th power;\n\n3.\\,(Inclusion–minimality)\\quad for every non-empty proper subset $I\\subset\\{1,\\dots ,r\\}$ the product\n   \\[\n   n\\cdot\\prod_{i\\in I}a_i\n   \\]\n   is \\emph{not} a perfect $k$-th power.\n\nAmong all $k$–admissible tuples for $n$ choose the lexicographically\nminimal one—first minimise $r$, and for that $r$ minimise $a_r$.\nDenote this distinguished tuple again by $A=(a_1,\\dots ,a_r)$ and set\n\\[\n\\Phi_k(n)\\;:=\\;(r,a_r)\\;\\in\\;\\mathbb N\\times\\mathbb N .\n\\]\n\n(a) Prove that for every $n\\in S_k$ at least one $k$–admissible tuple exists; hence $\\Phi_k(n)$ is well defined.\n\n(b) Show that for $k=2$ the map\n\\[\n\\Phi_2\\colon S_2\\longrightarrow\\mathbb N\\times\\mathbb N\n\\]\nis injective.\n\n(c) For every odd prime $k$ construct distinct integers\n$n,m\\in S_k$ with $\\Phi_k(n)=\\Phi_k(m)$, and deduce that\n$\\Phi_k$ fails to be injective whenever $k\\ge 3$.",
      "solution": "Throughout let $\\nu_p(\\,\\cdot\\,)$ denote the $p$–adic valuation and write \n\\[\n\\mathbf v(x)\\;:=\\;\\bigl(\\,\\nu_p(x)\\bmod k\\,\\bigr)_{p\\text{ prime}}\n          \\;\\in\\;\\bigoplus_{p}\\mathbb F_k\\;=:\\;V_k ,\n\\]\nwhere $\\mathbb F_k$ is the field with $k$ elements; addition in\n$V_k$ is component-wise modulo $k$.\n\n--------------------------------------------------------------------\nPart (a) – Existence of $k$–admissible tuples\n--------------------------------------------------------------------\nFactor $n$ as\n\\[\nn=t^{\\,k}s,\\qquad\ns=\\prod_{j=1}^{d}p_j^{e_j},\\quad 1\\le e_j\\le k-1 .\n\\]\n(The factor $s$ is the $k$-free part of $n$; since $n\\in S_k$ we have\n$s>1$.)  Choose pair-wise distinct primes \n\\[\nq_1,\\dots ,q_d>n,\\qquad q_i\\neq p_j\\ \\text{ for all }i,j,\n\\]\nand put \n\\[\na_j:=p_j^{\\,k-e_j}\\,q_j^{\\,k}\\qquad(1\\le j\\le d).\n\\]\n\n$\\bullet$ Order: $a_j>q_j>n$, and the $a_j$ are pair-wise distinct.\n\n$\\bullet$ Perfection: in the product $n\\cdot a_1\\dots a_d$ each\n$p_j$ occurs with exponent $e_j+(k-e_j)=k$ and each $q_j$ with\nexponent $k$, so the product is a perfect $k$-th power.\n\n$\\bullet$ Inclusion–minimality: omitting $a_j$ lowers the exponent\nof $p_j$ to $e_j\\not\\equiv 0\\pmod{k}$, destroying perfection.\n\nAfter increasing re-ordering we obtain a $k$–admissible tuple,\nso $\\Phi_k$ is well defined.\n\n--------------------------------------------------------------------\nPart (b) – Injectivity for $k=2$\n--------------------------------------------------------------------\nAssume\n\\[\nn,m\\in S_2,\\qquad n<m,\\qquad \\Phi_2(n)=\\Phi_2(m)=(r,M).\n\\]\nLet\n\\[\nA=(a_1,\\dots ,a_r),\\qquad\nB=(b_1,\\dots ,b_r)\n\\]\nbe the distinguished $2$–admissible tuples for $n$ and $m$,\nrespectively; hence $a_r=b_r=M$.\n\n--------------------------------------------------------------------\nCase $r=1$\n--------------------------------------------------------------------\nWrite $n=s\\,t^{\\,2}$ and $m=s'\\,u^{\\,2}$ with square-free\n$s,s'>1$ and integers $u>t\\ge 1$.  \nBecause the one-tuple $(M)$ works for \\emph{both} $n$ and $m$,\n$M$ is the smallest integer exceeding $n$ (respectively $m$)\nsuch that multiplication turns $n$ (respectively $m$) into a square, so\n\\[\nM=s\\,(t+1)^{2}=s'\\,(u+1)^{2}. \\tag{1}\n\\]\n\nSet $d=\\gcd(s,s')$ and write $s=d\\,s_0$, $s'=d\\,s_0'$ with\n$\\gcd(s_0,s_0')=1$.  Divide (1) by $d$:\n\\[\ns_0\\,(t+1)^{2}=s_0'\\,(u+1)^{2}. \\tag{2}\n\\]\n\n\\emph{Step 1: Showing $s_0=s_0'$.}  \nIf a prime $p$ divides $s_0$ but not $s_0'$, then comparing $p$-adic\nvaluations in (2) yields\n\\[\n1+2\\,\\nu_p(t+1)=2\\,\\nu_p(u+1),\n\\]\nwhich is impossible because the left side is odd while the right side\nis even.  By symmetry no prime divides exactly one of $s_0,s_0'$, hence\n$s_0=s_0'$.\n\n\\emph{Step 2: Showing $u=t$.}  \nWith $s_0=s_0'$ equation (2) reduces to $(t+1)^{2}=(u+1)^{2}$, so\n$u+1=t+1$ and therefore $u=t$.\n\nConsequently $m=s\\,u^{2}=s\\,t^{2}=n$, contradicting $n<m$.\nHence $r\\neq 1$.\n\n--------------------------------------------------------------------\nCase $r\\ge 2$\n--------------------------------------------------------------------\nIn the $\\mathbb F_2$–vector space $V_2$\n\\[\n\\mathbf v(n)+\\sum_{i=1}^{r}\\mathbf v(a_i)=\\mathbf 0,\\qquad\n\\mathbf v(m)+\\sum_{i=1}^{r}\\mathbf v(b_i)=\\mathbf 0.\\tag{$\\ast$}\n\\]\n\nLemma 1.  For any $2$–admissible tuple the vectors\n$\\mathbf v(a_1),\\dots ,\\mathbf v(a_r)$ are linearly independent\nin $V_2$.\n\nProof.  A non-trivial $\\mathbb F_2$–linear relation among them would\nallow deletion of at least one $a_j$ while preserving $(\\ast)$,\ncontradicting inclusion–minimality. \\hfill$\\square$\n\nHence\n\\[\nS:=\\{\\mathbf v(a_1),\\dots ,\\mathbf v(a_r)\\},\\qquad\nT:=\\{\\mathbf v(b_1),\\dots ,\\mathbf v(b_r)\\}\n\\]\nare bases of the same $r$–dimensional subspace\n$W=\\operatorname{span}(S)=\\operatorname{span}(T)$, and\n$\\mathbf u:=\\mathbf v(M)$ lies in $S\\cap T$.\n\nClaim 1. $T\\neq S$.\n\nIndeed, assume the contrary $T=S$.  Summing the two relations $(\\ast)$,\n\\[\n\\mathbf v(n)+\\mathbf v(m)=\\mathbf 0,\n\\]\nso every prime dividing $nm$ appears to an even total exponent,\nequivalently $m=nq^{\\,2}$ for some integer $q>1$.  Consider the single\ninteger $M':=nq^{\\,2}$.  Then $n<M'$ and $nM'=n^{\\,2}q^{\\,2}$ is a\nsquare, so $(M')$ is a $2$–admissible one-tuple for $n$, contradicting\n$r\\ge 2$.  Therefore $T\\neq S$ and an element\n\\[\n\\mathbf w\\in T\\setminus S\\qquad(\\mathbf w\\neq\\mathbf u)\n\\]\nexists.\n\nBy the exchange lemma\n\\[\nS':=(S\\setminus\\{\\mathbf u\\})\\cup\\{\\mathbf w\\}\n\\]\nis again a basis of $W$.  Choose $j<r$ with $\\mathbf v(b_j)=\\mathbf w$.\nBecause $\\mathbf w\\notin S$, $b_j\\notin\\{a_1,\\dots ,a_{r-1},a_r\\}$.\n\nDelete $M$ from $A$, insert $b_j$, and reorder increasingly to obtain\n$C=(c_1,\\dots ,c_r)$.  We verify that $C$ is $2$–admissible for $n$ with\n$c_r<M$, contradicting the choice of $A$.\n\n$\\bullet$ Order: $b_j<M$ by construction, so after re-ordering\n$c_r<M$ and the $c_i$ remain pair-wise distinct.\n\n$\\bullet$ Perfection: in $V_2$\n\\[\n\\mathbf v(n)+\\sum_{i=1}^{r}\\mathbf v(c_i)\n          =\\mathbf v(n)+\\sum_{\\mathbf x\\in S'}\\mathbf x\n          =\\mathbf v(n)+\\sum_{\\mathbf x\\in S}\\mathbf x\n          =\\mathbf 0,\n\\]\nso $n\\,c_1\\dots c_r$ is a square.\n\n$\\bullet$ Inclusion–minimality: Lemma 1 applied to the basis $S'$ shows\nits vectors are independent; hence no proper subproduct is a square.\n\nThus $C$ is a $2$–admissible tuple for $n$ whose last entry is\nstrictly smaller than $M$, a contradiction.  Therefore Case $r\\ge 2$\nis impossible, and $\\Phi_2$ is injective.\n\n--------------------------------------------------------------------\nPart (c) – Failure of injectivity for every odd prime $k$\n--------------------------------------------------------------------\nLet $k\\ge 3$ be prime.  Choose a prime $p>2^{\\,k}$ and set\n\\[\nn=p,\\qquad m=2^{\\,k}p.\n\\]\nBoth lie in $S_k$.  Define\n\\[\na_1:=b_1:=p^{\\,k-1}.\n\\]\nBecause $p^{\\,k-1}>2^{\\,k}p=m$ (since $p>2^{\\,k}$), the order\ncondition is satisfied for both $n$ and $m$.  Moreover\n\\[\nn\\,a_1      =p^{\\,k},\\qquad\nm\\,b_1      =(2p)^{\\,k},\n\\]\nso $(a_1)$ and $(b_1)$ are $k$–admissible tuples, whence\n\\[\n\\Phi_k(n)=\\Phi_k(m)=(1,p^{\\,k-1}).\\tag{3}\n\\]\n\nMinimality of $a_1=b_1$.  \nWe must show that no integer $\\ell$ with $n<\\ell<p^{\\,k-1}$\nis admissible for $n$ (and similarly for $m$).\n\n•  $p$–adic valuation.  \nFor $x\\in\\{n,m\\}$ we have $\\nu_p(x)=1\\not\\equiv 0\\pmod{k}$.\nIf $x\\ell$ is a perfect $k$-th power then\n\\[\n\\nu_p(\\ell)\\equiv -\\nu_p(x)\\equiv k-1\\pmod{k},\n\\]\nso $\\nu_p(\\ell)\\ge k-1$, hence $\\ell\\ge p^{\\,k-1}$.\n\n•  $2$–adic valuation (only relevant for $m$).  \nSince $\\nu_2(m)=k\\equiv 0\\pmod{k}$ already, the same congruence forces\n$\\nu_2(\\ell)\\equiv 0\\pmod{k}$.  The smallest possibility is\n$\\nu_2(\\ell)=0$, i.e. $\\ell$ is odd, which holds for $p^{\\,k-1}$ but\ndoes not yield any smaller admissible $\\ell$ because of the $p$–adic\nobstruction above.\n\nThus $p^{\\,k-1}$ is indeed the \\emph{smallest} admissible choice for\nboth $n$ and $m$, so the distinguished tuples are $(p^{\\,k-1})$,\nand (3) proves that $\\Phi_k$ is not injective for any odd prime $k$.\n\n--------------------------------------------------------------------\nConclusion\n--------------------------------------------------------------------\n(a) $k$–admissible tuples always exist;  \n(b) $\\Phi_2$ is injective;  \n(c) for every prime $k\\ge 3$ the map $\\Phi_k$ is not injective.\n\\hfill$\\blacksquare$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.825542",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher power setting.  The problem is no longer confined to\nperfect squares (k = 2) but to arbitrary k ≥ 2.  This forces the solver\nto handle exponents modulo k and to keep track of deficiency vectors\nfor every prime factor.\n\n2.  Minimal-subset requirement (condition (3)).  \nOne must produce admissible tuples that are *inclusion-minimal* and, in\nthe injectivity proof, exploit this minimality to rule out many\nsimpler counter-examples.  Controlling minimality greatly complicates\nboth the construction (Part (a)) and the structural argument\n(Part (b)).\n\n3.  Lexicographic invariant.  \nInstead of a single integer, the function Φₖ takes values in ℕ×ℕ with\nlexicographic order, so equal values impose two simultaneous equalities\n(ranks and maxima).  The injectivity proof has to consider both\ncoordinates and show that neither can be reduced without breaking\nadmissibility.\n\n4.  Additional algebraic tools.  \nThe solution consciously employs p-adic exponent vectors, modular\narithmetic of exponents, and multiset cancellations—techniques that go\nwell beyond the elementary argument sufficient for the original kernel\nvariant.\n\n5.  Generality.  \nEvery aspect (construction, cancellation, minimality) must work for\n*all* k ≥ 2, not merely for the quadratic case, adding several layers\nof bookkeeping and abstract reasoning.\n\nAltogether, these enhancements demand deeper number-theoretic insight,\nmore intricate constructions, and a substantially longer chain of\nlogical steps than the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $k\\ge 2$ be a fixed prime number and put \n\\[\nS_k \\;=\\;\\Bigl\\{\\,n\\in\\mathbb N \\;:\\; n\\text{ is \\emph{not} a perfect }k\\text{-th power}\\Bigr\\}.\n\\]\n\nFor $n\\in S_k$ call an ordered finite tuple \n\\[\nA=(a_1,\\dots ,a_r)\\qquad(r\\ge 1)\n\\]\n$k$–admissible for $n$ if  \n\n1.\\,(Order)\\; $n<a_1<a_2<\\dots <a_r$ and the $a_i$ are pairwise distinct;  \n\n2.\\,(Perfection)\\; the product $n\\cdot a_1a_2\\dots a_r$ is a perfect $k$-th power;  \n\n3.\\,(Inclusion–minimality)\\; for every non–empty proper subset $I\\subset\\{1,\\dots ,r\\}$ the product\n   \\[\n   n\\cdot\\prod_{i\\in I}a_i\n   \\]\n   is \\emph{not} a perfect $k$-th power.\n\nAmong all $k$–admissible tuples for $n$ choose the lexicographically\nminimal one—first minimise $r$, and for that $r$ minimise $a_r$.\nDenote this distinguished tuple again by $A=(a_1,\\dots ,a_r)$ and set \n\\[\n\\Phi_k(n)\\;:=\\;(r,a_r)\\;\\in\\;\\mathbb N\\times\\mathbb N .\n\\]\n\n(a) Prove that for every $n\\in S_k$ at least one $k$–admissible tuple exists; hence $\\Phi_k(n)$ is well defined.\n\n(b) Show that for $k=2$ the map\n\\[\n\\Phi_2\\colon S_2\\longrightarrow\\mathbb N\\times\\mathbb N\n\\]\nis injective.\n\n(c) For every odd prime $k$ construct distinct integers\n$n,m\\in S_k$ with $\\Phi_k(n)=\\Phi_k(m)$, and deduce that\n$\\Phi_k$ fails to be injective whenever $k\\ge 3$.",
      "solution": "Throughout let $\\nu_p(\\,\\cdot\\,)$ denote the $p$–adic valuation and write \n\\[\n\\mathbf v(x)\\;:=\\;\\bigl(\\,\\nu_p(x)\\bmod k\\,\\bigr)_{p\\text{ prime}}\n          \\;\\in\\;\\bigoplus_{p}\\mathbb F_k\\;=:\\;V_k ,\n\\]\nwhere $\\mathbb F_k$ is the field with $k$ elements; addition in\n$V_k$ is componentwise modulo $k$.\n\n--------------------------------------------------------------------\nPart (a) – Existence of $k$–admissible tuples\n--------------------------------------------------------------------\nFactor $n$ as\n\\[\nn=t^{\\,k}s,\\qquad\ns=\\prod_{j=1}^{d}p_j^{e_j},\\quad 1\\le e_j\\le k-1 .\n\\]\n(The factor $s$ is the $k$–free part of $n$; since $n\\in S_k$ we have\n$s>1$.)  Choose pairwise distinct primes  \n\\[\nq_1,\\dots ,q_d>n,\\qquad q_i\\neq p_j\\ \\text{ for all }i,j,\n\\]\nand put  \n\\[\na_j:=p_j^{\\,k-e_j}\\,q_j^{\\,k}\\qquad(1\\le j\\le d).\n\\]\n\n$\\bullet$ Order: $a_j>q_j>n$, and the $a_j$ are pairwise distinct.\n\n$\\bullet$ Perfection: in the product $n\\cdot a_1\\dots a_d$ each\n$p_j$ occurs with exponent $e_j+(k-e_j)=k$ and each $q_j$ with\nexponent $k$, so the product is a perfect $k$–th power.\n\n$\\bullet$ Inclusion–minimality: omitting $a_j$ lowers the exponent\nof $p_j$ to $e_j\\not\\equiv 0\\pmod{k}$, destroying perfection.\n\nAfter increasing re–ordering we obtain a $k$–admissible tuple,\nso $\\Phi_k$ is well defined.\n\n--------------------------------------------------------------------\nPart (b) – Injectivity for $k=2$\n--------------------------------------------------------------------\nAssume\n\\[\nn,m\\in S_2,\\qquad n<m,\\qquad \\Phi_2(n)=\\Phi_2(m)=(r,M).\n\\]\nLet\n\\[\nA=(a_1,\\dots ,a_r),\\qquad\nB=(b_1,\\dots ,b_r)\n\\]\nbe the distinguished $2$–admissible tuples for $n$ and $m$,\nrespectively; hence $a_r=b_r=M$.\n\n--------------------------------------------------------------------\nCase $r=1$\n--------------------------------------------------------------------\nWrite $n=s\\,t^{\\,2}$, $m=s'\\,u^{\\,2}$ with square–free\n$s,s'>1$ and integers $u>t\\ge 1$.  \nBecause the one–tuple $(M)$ works for \\emph{both} $n$ and $m$,\n$M$ is the smallest integer exceeding $n$ (respectively $m$)\nsuch that multiplication turns $n$ (respectively $m$) into a square.\nNecessarily\n\\[\nM=s\\,(t+1)^{2}\\quad\\text{and}\\quad M=s'\\,(u+1)^{2},\n\\]\nfor if $\\ell\\le t$ then $s\\,\\ell^{2}\\le n<M$, contradicting the\nminimality of $M$, and similarly for $m$.\n\nDividing the two equalities gives\n\\[\n\\frac{s}{s'}=\\biggl(\\frac{u+1}{t+1}\\biggr)^{\\!2}.\n\\]\nThe left–hand side is square–free, hence equals $1$; thus $s=s'$ and\n$(u+1)^{2}=(t+1)^{2}$, whence $u=t$.  Consequently $m=s\\,u^{2}=s\\,t^{2}=n$,\ncontradicting $n<m$.  Therefore $r\\neq 1$.\n\n--------------------------------------------------------------------\nCase $r\\ge 2$\n--------------------------------------------------------------------\nIn the $\\mathbb F_2$–vector space $V_2$\n\\[\n\\mathbf v(n)+\\sum_{i=1}^{r}\\mathbf v(a_i)=\\mathbf 0,\\qquad\n\\mathbf v(m)+\\sum_{i=1}^{r}\\mathbf v(b_i)=\\mathbf 0.\\tag{$\\ast$}\n\\]\n\nLemma 1.  For any $2$–admissible tuple the vectors\n$\\mathbf v(a_1),\\dots ,\\mathbf v(a_r)$ are linearly independent\nin $V_2$.\n\nProof.  A non–trivial $\\mathbb F_2$–linear relation among them would\nallow deletion of at least one $a_j$ while preserving $(\\ast)$,\ncontradicting inclusion–minimality. \\hfill$\\square$\n\nHence\n\\[\nS:=\\{\\mathbf v(a_1),\\dots ,\\mathbf v(a_r)\\},\\qquad\nT:=\\{\\mathbf v(b_1),\\dots ,\\mathbf v(b_r)\\}\n\\]\nare bases of the same $r$–dimensional subspace\n$W=\\operatorname{span}(S)=\\operatorname{span}(T)$, and\n$\\mathbf u:=\\mathbf v(M)$ lies in $S\\cap T$.\n\nClaim 1. $T\\neq S$.\n\nIndeed, assume the contrary $T=S$.  Summing the two relations $(\\ast)$,\n\\[\n\\mathbf v(n)+\\mathbf v(m)=\\mathbf 0,\n\\]\nso every prime dividing $nm$ appears to an even total exponent,\nequivalently $m=nq^{\\,2}$ for some integer $q>1$.  Consider the single\ninteger $M':=nq^{\\,2}$.  Then $n<M'$ and $nM'=n^{\\,2}q^{\\,2}$ is a\nsquare, so $(M')$ is a $2$–admissible one–tuple for $n$, contradicting\n$r\\ge 2$.  Therefore $T\\neq S$ and an element\n\\[\n\\mathbf w\\in T\\setminus S\\qquad(\\mathbf w\\neq\\mathbf u)\n\\]\nexists.\n\nBy the exchange lemma\n\\[\nS':=(S\\setminus\\{\\mathbf u\\})\\cup\\{\\mathbf w\\}\n\\]\nis again a basis of $W$.  Choose $j<r$ with $\\mathbf v(b_j)=\\mathbf w$.\nBecause $\\mathbf w\\notin S$, $b_j\\notin\\{a_1,\\dots ,a_{r-1},a_r\\}$.\n\nDelete $M$ from $A$, insert $b_j$, and reorder increasingly to obtain\n$C=(c_1,\\dots ,c_r)$.  We verify that $C$ is $2$–admissible for $n$ with\n$c_r<M$, contradicting the choice of $A$.\n\n$\\bullet$ Order: $b_j<M$ by construction, so after re–ordering\n$c_r<M$ and the $c_i$ remain pairwise distinct.\n\n$\\bullet$ Perfection: in $V_2$\n\\[\n\\mathbf v(n)+\\sum_{i=1}^{r}\\mathbf v(c_i)\n          =\\mathbf v(n)+\\sum_{\\mathbf x\\in S'}\\mathbf x\n          =\\mathbf v(n)+\\sum_{\\mathbf x\\in S}\\mathbf x\n          =\\mathbf 0,\n\\]\nso $n\\,c_1\\dots c_r$ is a square.\n\n$\\bullet$ Inclusion–minimality: Lemma 1 applied to the basis $S'$ shows\nits vectors are independent; hence no proper subproduct is a square.\n\nThus $C$ is a $2$–admissible tuple for $n$ whose last entry is\nstrictly smaller than $M$, a contradiction.  Therefore Case $r\\ge 2$\nis impossible, and $\\Phi_2$ is injective.\n\n--------------------------------------------------------------------\nPart (c) – Failure of injectivity for every odd prime $k$\n--------------------------------------------------------------------\nLet $k\\ge 3$ be prime.  Choose a prime $p>2^{\\,k}$ and set\n\\[\nn=p,\\qquad m=2^{\\,k}p.\n\\]\nBoth lie in $S_k$.  Define\n\\[\na_1:=b_1:=p^{\\,k-1}.\n\\]\nBecause $p^{\\,k-1}>2^{\\,k}p=m$ (since $p>2^{\\,k}$ and $k\\ge 3$),\nthe order condition is satisfied for both $n$ and $m$.  Moreover\n\\[\nn\\,a_1      =p^{\\,k},\\qquad\nm\\,b_1      =(2p)^{\\,k},\n\\]\nso $(a_1)$ and $(b_1)$ are $k$–admissible tuples, whence\n\\[\n\\Phi_k(n)=\\Phi_k(m)=(1,p^{\\,k-1})\\quad\\text{and }n\\neq m .\n\\]\nThus $\\Phi_k$ is not injective for any odd prime $k$.\n\n--------------------------------------------------------------------\nConclusion\n--------------------------------------------------------------------\n(a) $k$–admissible tuples always exist;  \n(b) $\\Phi_2$ is injective;  \n(c) for every prime $k\\ge 3$ the map $\\Phi_k$ is not injective.\n\\hfill$\\blacksquare$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.631332",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher power setting.  The problem is no longer confined to\nperfect squares (k = 2) but to arbitrary k ≥ 2.  This forces the solver\nto handle exponents modulo k and to keep track of deficiency vectors\nfor every prime factor.\n\n2.  Minimal-subset requirement (condition (3)).  \nOne must produce admissible tuples that are *inclusion-minimal* and, in\nthe injectivity proof, exploit this minimality to rule out many\nsimpler counter-examples.  Controlling minimality greatly complicates\nboth the construction (Part (a)) and the structural argument\n(Part (b)).\n\n3.  Lexicographic invariant.  \nInstead of a single integer, the function Φₖ takes values in ℕ×ℕ with\nlexicographic order, so equal values impose two simultaneous equalities\n(ranks and maxima).  The injectivity proof has to consider both\ncoordinates and show that neither can be reduced without breaking\nadmissibility.\n\n4.  Additional algebraic tools.  \nThe solution consciously employs p-adic exponent vectors, modular\narithmetic of exponents, and multiset cancellations—techniques that go\nwell beyond the elementary argument sufficient for the original kernel\nvariant.\n\n5.  Generality.  \nEvery aspect (construction, cancellation, minimality) must work for\n*all* k ≥ 2, not merely for the quadratic case, adding several layers\nof bookkeeping and abstract reasoning.\n\nAltogether, these enhancements demand deeper number-theoretic insight,\nmore intricate constructions, and a substantially longer chain of\nlogical steps than the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}