Erdős Problem #978

Erdős Problem 978 --- collection record

PROVENANCE AND STATUS

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_5.2/978.tex.

REUSED SOURCE 1

1) FORMAL RESTATEMENT

Let $f\in\mathbb Z[x]$ be irreducible of degree $k>2$. An integer $N\ne 0$ is called $t$-power-free if there is no prime $p$ with $p^t\mid N$. The questions are: (a) for $k>2$ with $k\ne 2^\ell$, does the set $\{n\in\mathbb N: f(n)\text{ is }(k-1)\text{-power-free}\}$ have positive (natural) density? (b) are there infinitely many $n$ such that $f(n)$ is $(k-2)$-power-free? In particular (taking $f(n)=n^4+2$ and $k=4$), are there infinitely many $n$ for which $n^4+2$ is squarefree?

Conventions/edge cases: - We interpret $t$-power-free for negative values via $|f(n)|$. - The density in (a) is $\lim_{x\to\infty}\frac1x\#\{1\le n\le x: f(n)\text{ is }(k-1)\text{-power-free}\}$ when the limit exists.

2) QUICK LITERATURE/CONTEXT CHECK

The provided statement already records several known results (Erdős, Hooley, Heath-Brown, Browning). A quick web search $2026-01-16$ did not surface an explicit resolution of the specific squarefree question for $n^4+2$ beyond what is written in the problem statement, so I treat that question as open here.

3) ATTACK PLAN

Proof track: one expects to use a power-free (squarefree) sieve, reducing the problem to controlling solutions of congruences $f(n)\equiv 0\pmod{p^{k-2}}$ for many primes $p$, and then estimating the remainder term. Disproof track: look for a local obstruction (a prime $p$ such that $p^{k-2}\mid f(n)$ for all $n$), which would force only finitely many power-free values.

In this write-up I give elementary local computations for the special polynomial $n^4+2$ and a general lemma for the simplest local obstruction $p\mid f(n)$ for all $n$.

4) WORK

FAST REALITY CHECK (for $f(n)=n^4+2$; exact computations)

Count of $1\le n\le N$ for which $n^4+2$ is squarefree: \[ \begin{array}{c|c|c} N & \#\{1\le n\le N: n^4+2\text{ squarefree}\} & \text{fraction}\\\hline 10 & 8 & 0.8\\ 50 & 38 & 0.76\\ 100 & 75 & 0.75\\ 500 & 379 & 0.758\\ 1000 & 759 & 0.759\\ 2000 & 1514 & 0.757\\ 5000 & 3787 & 0.7574 \end{array} \] Among $1\le n\le 100$, the first few nonsquarefree values occur at $n=2$ ($n^4+2=18=2\cdot 3^2$), $n=7$ ($2403=3^3\cdot 89$), $n=11$ ($14643=3^2\cdot 1627$).

Lemma 4.1 (a basic local obstruction criterion mod $p$). Let $p$ be prime and let $\overline f\in\mathbb F_p[x]$ be the reduction of $f$ modulo $p$. Then the following are equivalent: (i) $p\mid f(n)$ for every integer $n$. (ii) $\overline f(a)=0$ for every $a\in\mathbb F_p$. (iii) $x^p-x$ divides $\overline f(x)$ in $\mathbb F_p[x]$.

Proof. (i)$\Rightarrow$(ii): reduce the congruence $f(n)\equiv 0\pmod p$ modulo $p$. (ii)$\Leftrightarrow$(iii): in $\mathbb F_p[x]$ we have the factorization $x^p-x=\prod_{a\in\mathbb F_p}(x-a)$, so a polynomial vanishes on every element of $\mathbb F_p$ iff it has every $a\in\mathbb F_p$ as a root, equivalently iff it is divisible by $\prod_{a\in\mathbb F_p}(x-a)=x^p-x$. (ii)$\Rightarrow$(i): if $\overline f$ vanishes on all residues modulo $p$, then $f(n)\equiv 0\pmod p$ for every integer $n$. $\square$

Lemma 4.2 (exact $3$-adic divisibility pattern for $n^4+2$ at level $3^2$). For every integer $n$, \[ 9\mid (n^4+2)\quad\Longleftrightarrow\quad n\equiv 2\text{ or }7\pmod 9. \] Moreover, $3\mid(n^4+2)$ if and only if $3\nmid n$.

Proof. Modulo $3$, one has $n^4\equiv 0$ if $3\mid n$ and $n^4\equiv 1$ if $3\nmid n$. Thus $n^4+2\equiv 2\pmod 3$ when $3\mid n$ and $n^4+2\equiv 0\pmod 3$ when $3\nmid n$, proving the second claim. For the $\bmod\ 9$ statement, compute $(r^4+2)\bmod 9$ for residues $r=0,1,\dots,8$: \[ \begin{array}{c|ccccccccc} r & 0&1&2&3&4&5&6&7&8\\\hline r^4+2\pmod 9 & 2&3&0&2&6&6&2&0&3 \end{array} \] So $9\mid (n^4+2)$ exactly for $n\equiv 2,7\pmod 9$. $\square$

Lemma 4.3 (no fixed square divisor for $n^4+2$). There is no prime $p$ such that $p^2\mid n^4+2$ for all integers $n$.

Proof. If $p^2\mid n^4+2$ for all $n$, then in particular $p^2\mid 0^4+2=2$, hence $p=2$. But for $n=1$ we have $1^4+2=3$, which is not divisible by $4$. Contradiction. $\square$

5) VERIFICATION

- Lemma 4.1 is purely over $\mathbb F_p$ and uses the standard identity $x^p-x=\prod_{a\in\mathbb F_p}(x-a)$. - Lemma 4.2 was verified by an explicit residue computation modulo $9$. - Lemma 4.3 uses only evaluations at $n=0$ and $n=1$, hence is robust.

6) FINAL

UNRESOLVED (i) Strongest proved partial result: for the specific polynomial $n^4+2$, the exact criterion for divisibility by $9$ is $n\equiv 2,7\pmod 9$ (Lemma 4.2), and there is no fixed prime square dividing all values (Lemma 4.3). Computations up to $N=5000$ show $3787$ squarefree values among the first $5000$ inputs (table above). (ii) First gap (crisp): prove that the set $\{n\ge 1: n^4+2\text{ is squarefree}\}$ is infinite (or give a counterexample by producing a prime square dividing $n^4+2$ for all sufficiently large $n$, which seems incompatible with Lemma 4.3 but could in principle occur intermittently). (iii) Top 3 next moves: 1. For each prime $p$, bound the number of residue classes $n\bmod p^2$ with $n^4\equiv -2\pmod{p^2}$ uniformly and sum the resulting local densities to set up a squarefree sieve. 2. Control the remainder term in the sieve for $n^4+2$ by estimating correlations of the events $p^2\mid(n^4+2)$ for different primes $p$. 3. Computation: for increasing $N$, factor $n^4+2$ and record which prime squares occur, to guess the main obstructions (e.g. the prominent $3^2$ obstruction coming from Lemma 4.2). (iv) Minimal counterexample structure: if $n^4+2$ were squarefree only finitely often, then beyond some $N_0$ every value would be divisible by $p(n)^2$ for some prime $p(n)$ depending on $n$; one would expect a structured concentration of solutions to $n^4\equiv -2\pmod{p^2}$ for many primes $p$, contrary to the heuristic that such congruences have $O(1)$ solutions modulo $p^2$.

COMPLETION ESTIMATE

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

1) FORMAL RESTATEMENT

Let $f\in\mathbb Z[x]$ be irreducible of degree $k>2$. An integer $N\ne 0$ is called $t$-power-free if there is no prime $p$ with $p^t\mid N$. The questions are: (a) for $k>2$ with $k\ne 2^\ell$, does the set $\{n\in\mathbb N: f(n)\text{ is }(k-1)\text{-power-free}\}$ have positive (natural) density? (b) are there infinitely many $n$ such that $f(n)$ is $(k-2)$-power-free? In particular (taking $f(n)=n^4+2$ and $k=4$), are there infinitely many $n$ for which $n^4+2$ is squarefree?

Conventions/edge cases: - We interpret $t$-power-free for negative values via $|f(n)|$. - The density in (a) is $\lim_{x\to\infty}\frac1x\#\{1\le n\le x: f(n)\text{ is }(k-1)\text{-power-free}\}$ when the limit exists.

2) QUICK LITERATURE/CONTEXT CHECK

The provided statement already records several known results (Erdős, Hooley, Heath-Brown, Browning). A quick web search $2026-01-16$ did not surface an explicit resolution of the specific squarefree question for $n^4+2$ beyond what is written in the problem statement, so I treat that question as open here.

3) ATTACK PLAN

Proof track: one expects to use a power-free (squarefree) sieve, reducing the problem to controlling solutions of congruences $f(n)\equiv 0\pmod{p^{k-2}}$ for many primes $p$, and then estimating the remainder term. Disproof track: look for a local obstruction (a prime $p$ such that $p^{k-2}\mid f(n)$ for all $n$), which would force only finitely many power-free values.

In this write-up I give elementary local computations for the special polynomial $n^4+2$ and a general lemma for the simplest local obstruction $p\mid f(n)$ for all $n$.

4) WORK

FAST REALITY CHECK (for $f(n)=n^4+2$; exact computations)

Count of $1\le n\le N$ for which $n^4+2$ is squarefree: \[ \begin{array}{c|c|c} N & \#\{1\le n\le N: n^4+2\text{ squarefree}\} & \text{fraction}\\\hline 10 & 8 & 0.8\\ 50 & 38 & 0.76\\ 100 & 75 & 0.75\\ 500 & 379 & 0.758\\ 1000 & 759 & 0.759\\ 2000 & 1514 & 0.757\\ 5000 & 3787 & 0.7574 \end{array} \] Among $1\le n\le 100$, the first few nonsquarefree values occur at $n=2$ ($n^4+2=18=2\cdot 3^2$), $n=7$ ($2403=3^3\cdot 89$), $n=11$ ($14643=3^2\cdot 1627$).

Lemma 4.1 (a basic local obstruction criterion mod $p$). Let $p$ be prime and let $\overline f\in\mathbb F_p[x]$ be the reduction of $f$ modulo $p$. Then the following are equivalent: (i) $p\mid f(n)$ for every integer $n$. (ii) $\overline f(a)=0$ for every $a\in\mathbb F_p$. (iii) $x^p-x$ divides $\overline f(x)$ in $\mathbb F_p[x]$.

Proof. (i)$\Rightarrow$(ii): reduce the congruence $f(n)\equiv 0\pmod p$ modulo $p$. (ii)$\Leftrightarrow$(iii): in $\mathbb F_p[x]$ we have the factorization $x^p-x=\prod_{a\in\mathbb F_p}(x-a)$, so a polynomial vanishes on every element of $\mathbb F_p$ iff it has every $a\in\mathbb F_p$ as a root, equivalently iff it is divisible by $\prod_{a\in\mathbb F_p}(x-a)=x^p-x$. (ii)$\Rightarrow$(i): if $\overline f$ vanishes on all residues modulo $p$, then $f(n)\equiv 0\pmod p$ for every integer $n$. $\square$

Lemma 4.2 (exact $3$-adic divisibility pattern for $n^4+2$ at level $3^2$). For every integer $n$, \[ 9\mid (n^4+2)\quad\Longleftrightarrow\quad n\equiv 2\text{ or }7\pmod 9. \] Moreover, $3\mid(n^4+2)$ if and only if $3\nmid n$.

Proof. Modulo $3$, one has $n^4\equiv 0$ if $3\mid n$ and $n^4\equiv 1$ if $3\nmid n$. Thus $n^4+2\equiv 2\pmod 3$ when $3\mid n$ and $n^4+2\equiv 0\pmod 3$ when $3\nmid n$, proving the second claim. For the $\bmod\ 9$ statement, compute $(r^4+2)\bmod 9$ for residues $r=0,1,\dots,8$: \[ \begin{array}{c|ccccccccc} r & 0&1&2&3&4&5&6&7&8\\\hline r^4+2\pmod 9 & 2&3&0&2&6&6&2&0&3 \end{array} \] So $9\mid (n^4+2)$ exactly for $n\equiv 2,7\pmod 9$. $\square$

Lemma 4.3 (no fixed square divisor for $n^4+2$). There is no prime $p$ such that $p^2\mid n^4+2$ for all integers $n$.

Proof. If $p^2\mid n^4+2$ for all $n$, then in particular $p^2\mid 0^4+2=2$, hence $p=2$. But for $n=1$ we have $1^4+2=3$, which is not divisible by $4$. Contradiction. $\square$

5) VERIFICATION

- Lemma 4.1 is purely over $\mathbb F_p$ and uses the standard identity $x^p-x=\prod_{a\in\mathbb F_p}(x-a)$. - Lemma 4.2 was verified by an explicit residue computation modulo $9$. - Lemma 4.3 uses only evaluations at $n=0$ and $n=1$, hence is robust.

6) FINAL

UNRESOLVED (i) Strongest proved partial result: for the specific polynomial $n^4+2$, the exact criterion for divisibility by $9$ is $n\equiv 2,7\pmod 9$ (Lemma 4.2), and there is no fixed prime square dividing all values (Lemma 4.3). Computations up to $N=5000$ show $3787$ squarefree values among the first $5000$ inputs (table above). (ii) First gap (crisp): prove that the set $\{n\ge 1: n^4+2\text{ is squarefree}\}$ is infinite (or give a counterexample by producing a prime square dividing $n^4+2$ for all sufficiently large $n$, which seems incompatible with Lemma 4.3 but could in principle occur intermittently). (iii) Top 3 next moves: 1. For each prime $p$, bound the number of residue classes $n\bmod p^2$ with $n^4\equiv -2\pmod{p^2}$ uniformly and sum the resulting local densities to set up a squarefree sieve. 2. Control the remainder term in the sieve for $n^4+2$ by estimating correlations of the events $p^2\mid(n^4+2)$ for different primes $p$. 3. Computation: for increasing $N$, factor $n^4+2$ and record which prime squares occur, to guess the main obstructions (e.g. the prominent $3^2$ obstruction coming from Lemma 4.2). (iv) Minimal counterexample structure: if $n^4+2$ were squarefree only finitely often, then beyond some $N_0$ every value would be divisible by $p(n)^2$ for some prime $p(n)$ depending on $n$; one would expect a structured concentration of solutions to $n^4\equiv -2\pmod{p^2}$ for many primes $p$, contrary to the heuristic that such congruences have $O(1)$ solutions modulo $p^2$.