Reused writeup on Math-6030 (gpt pro 5.2)

Erdős Problem 976 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/976

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/976.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/976.tex.

1) FORMAL RESTATEMENT

Let $f\in\mathbb Z[x]$ be irreducible over $\mathbb Z$ and have degree $d\ge 2$. For an integer $N\ne 0$, write $P^+(N)$ for the largest prime divisor of $|N|$. For $n\ge 1$ define \[ F_f(n)\;:=\;\max_{1\le m\le n} P^+\left(f(m)\right). \] Equivalently, \[ F_f(n)\;=\; P^+\left(\prod_{m=1}^n f(m)\right), \] where prime divisors are taken of the absolute value. The problem asks for estimates of $F_f(n)$ as $n\to\infty$, and in particular asks whether there exists $c>0$ (depending on $f$) such that $F_f(n)\gg n^{1+c}$, or even $F_f(n)\gg n^d$.

Edge cases/conventions: - Since $f$ is irreducible of degree $\ge 2$, $f$ has no integer roots, hence $f(m)\ne 0$ for all integers $m$. - $f(m)$ may be negative; prime divisors are those of $|f(m)|$.

2) QUICK LITERATURE/CONTEXT CHECK

The provided problem statement already lists several lower bounds and references. A quick web search $2026-01-16$ did not surface an explicit resolution of the questions beyond what is already stated there, so I treat the problem as open in the form asked.

3) ATTACK PLAN

Proof track (high-level): if all prime factors of $\prod_{m\le n} f(m)$ were $\le y$, then the product would be $y$-smooth; to contradict this one needs upper bounds on the total $p$-adic valuation contributed by small primes. This typically requires delicate control of the number of solutions of $f(x)\equiv 0\pmod{p^a}$ for many primes and exponents.

Disproof track (high-level): try to construct an irreducible $f$ for which $f(1),\dots,f(n)$ are unusually smooth for infinitely many $n$.

In this write-up I only establish elementary structural lemmas and provide a fast computational sanity check for sample polynomials.

4) WORK

FAST REALITY CHECK (sample polynomials; exact computations)

For $f(x)=x^2+1$ (degree $2$): \[ \begin{array}{c|c|c} n & F_f(n) & \text{witness }m\text{ with }P^+(f(m))=F_f(n)\\\hline 5 & 17 & m=4,\ f(4)=17\\ 10 & 101 & m=10,\ f(10)=101\\ 20 & 401 & m=20,\ f(20)=401\\ 50 & 1601 & m=40,\ f(40)=1601\\ 100 & 8837 & m=94,\ f(94)=8837\\ 200 & 33857 & m=184,\ f(184)=33857 \end{array} \]

For $f(x)=x^3+2$ (degree $3$): \[ \begin{array}{c|c|c} n & F_f(n) & \text{witness }m\\\hline 5 & 127 & m=5,\ f(5)=127\\ 10 & 257 & m=8,\ f(8)=514=2\cdot 257\\ 20 & 4001 & m=20,\ f(20)=8002=2\cdot 4001\\ 50 & 91127 & m=45,\ f(45)=91127\\ 100 & 571789 & m=83,\ f(83)=571789\\ 200 & 6751271 & m=189,\ f(189)=6751271 \end{array} \] These samples are consistent with (but do not prove) growth on the order of $n^d$.

Lemma 4.1 (equivalence of definitions). For every $n\ge 1$, \[ \max_{1\le m\le n} P^+\left(f(m)\right) \;=\; P^+\left(\prod_{m=1}^n f(m)\right). \]

Proof. Let $M:=\prod_{m=1}^n f(m)$. Every prime divisor of some $f(m)$ divides $M$, so $\max_{m\le n}P^+(f(m))\le P^+(M)$. Conversely, any prime divisor of $M$ divides some factor $f(m)$, so $P^+(M)\le \max_{m\le n}P^+(f(m))$. $\square$

Lemma 4.2 (polynomial growth gives a trivial upper bound). There exists a constant $C_f>0$ such that for all $n\ge 1$, \[ F_f(n)\le C_f\, n^d. \]

Proof. Write $f(x)=\sum_{i=0}^d a_i x^i$ with $a_i\in\mathbb Z$ and $a_d\ne 0$. For $1\le m\le n$, \[ |f(m)|\le \sum_{i=0}^d |a_i|\, |m|^i\le \left(\sum_{i=0}^d |a_i|\right) n^d. \] Let $C_f:=\sum_{i=0}^d |a_i|$. Any prime divisor of $f(m)$ is at most $|f(m)|$, hence $P^+(f(m))\le C_f n^d$ for each $m\le n$, and taking a maximum gives the claim. $\square$

Lemma 4.3 (bounded gcd of consecutive values via a Bezout identity). Let $g(x):=f(x+1)\in\mathbb Z[x]$. Then $f$ and $g$ are coprime in $\mathbb Q[x]$, hence there exist polynomials $A,B\in\mathbb Z[x]$ and a nonzero integer $R$ such that \[ A(x)f(x)+B(x)f(x+1)=R\quad\text{in }\mathbb Z[x]. \] Consequently, for every integer $n$, \[ \gcd\left(f(n),f(n+1)\right)\mid R. \] In particular, if a prime $p$ divides both $f(n)$ and $f(n+1)$ for some $n$, then $p\mid R$.

Proof. First, $f$ and $f(x+1)$ are not associates in $\mathbb Q[x]$: if $f(x+1)=c f(x)$ for some constant $c\in\mathbb Q^\times$, comparing leading coefficients forces $c=1$, hence $f(x+1)=f(x)$, impossible for a nonconstant polynomial. Therefore $\gcd(f,g)=1$ in $\mathbb Q[x]$. So there exist $\alpha,\beta\in\mathbb Q[x]$ such that $\alpha(x)f(x)+\beta(x)g(x)=1$. Clearing denominators yields $A,B\in\mathbb Z[x]$ and $R\in\mathbb Z\setminus\{0\}$ with $A f + B g = R$. Evaluating at an integer $n$ gives $A(n)f(n)+B(n)f(n+1)=R$, hence any common divisor of $f(n)$ and $f(n+1)$ divides $R$. $\square$

Lemma 4.4 (root-count bound mod $p$ gives a basic divisor-count bound). Assume $f\not\equiv 0\pmod p$ in $(\mathbb Z/p\mathbb Z)[x]$. Then the congruence $f(m)\equiv 0\pmod p$ has at most $d$ solutions modulo $p$. Consequently, for every $n\ge 1$, \[ \#\{1\le m\le n: p\mid f(m)\}\le d\,\left(\lfloor\frac{n}{p}\rfloor+1\right). \]

Proof. Over the field $\mathbb F_p$, a nonzero polynomial of degree $d$ has at most $d$ roots. Let $S\subset\{0,1,\dots,p-1\}$ be the set of residue classes $r$ with $f(r)\equiv 0\pmod p$; then $|S|\le d$. For each $r\in S$, the integers $m\in[1,n]$ with $m\equiv r\pmod p$ form an arithmetic progression of size at most $\lfloor n/p\rfloor+1$. Summing over $r\in S$ gives the stated bound. $\square$

5) VERIFICATION

- The upper bound (Lemma 4.2) is a direct size bound and is valid even when $f(m)$ changes sign. - Lemma 4.3: the only subtle point is that $f(x)$ and $f(x+1)$ are coprime in $\mathbb Q[x]$. If they had a common factor in $\mathbb Q[x]$, then since $f$ is irreducible, $f$ itself would divide $f(x+1)$, forcing $f(x+1)=f(x)$ and contradicting nonconstancy. - Lemma 4.4 requires $f\not\equiv 0\pmod p$; since $f$ is irreducible, it is primitive, but this does not prevent $f\equiv 0\pmod p$ for some $p$ (it would require every coefficient divisible by $p$, which cannot happen for a primitive polynomial). Thus for irreducible $f\in\mathbb Z[x]$, the hypothesis holds for every prime $p$.

6) FINAL

UNRESOLVED (i) Strongest proved partial result: elementary structure bounds include the trivial upper bound $F_f(n)\le C_f n^d$ (Lemma 4.2), a uniform bound on $\gcd(f(n),f(n+1))$ (Lemma 4.3), and a uniform bound on the count of $m\le n$ for which a given prime $p$ divides $f(m)$ (Lemma 4.4). Exact computations for sample irreducible polynomials of degrees $2$ and $3$ up to $n=200$ are recorded above. (ii) First gap (crisp): prove any superlinear lower bound of the form $F_f(n)\ge c\,n^{1+\varepsilon}$ (or even $\ge c\,n\log n$) for a general irreducible $f\in\mathbb Z[x]$ by an argument that controls contributions of small primes to $\prod_{m\le n} f(m)$. (iii) Top 3 next moves: 1. Prove a uniform bound (in terms of $d$) on $\#\{x\bmod p^a: f(x)\equiv 0\pmod{p^a}\}$ for most primes $p$ and all $a\ge 1$, and turn it into a bound on $\sum_{m\le n} v_p(f(m))$. 2. Combine such $p$-adic bounds with a smooth-number estimate to show that assuming all prime factors are $\le y$ forces $\log\prod_{m\le n}|f(m)|\ll n\log y$, contradicting the $\asymp d n\log n$ growth. 3. Computationally: for additional polynomials $f$ (varying degrees and leading coefficients), compute $F_f(n)$ and track the ratio $F_f(n)/n^d$ and the location of the maximizing $m$ to look for patterns that might inform the $p$-adic counting step. (iv) Minimal counterexample structure: a putative obstruction to $F_f(n)\gg n^{1+\varepsilon}$ would be an irreducible polynomial $f$ together with an infinite sequence $n_j\to\infty$ such that all values $f(1),\dots,f(n_j)$ are unusually $y_j$-smooth with $y_j\le n_j^{1+\varepsilon}$, i.e. the set of prime divisors of $\prod_{m\le n_j} f(m)$ stays below $n_j^{1+\varepsilon}$.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.