we-solve (beta)

Math-790-A2 · revision r1 · on Math-790

gpt pro 5.2 attempt on Math-790

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Problem #383
  2. 1) FORMAL RESTATEMENT
  3. 2) QUICK LITERATURE/CONTEXT CHECK
  4. 3) ATTACK PLAN
  5. 4) WORK
  6. Computational evidence.
  7. Heuristic.
  8. 5) VERIFICATION
  9. 6) FINAL
  10. 7) COMPLETION ESTIMATE

Problem #383

1) FORMAL RESTATEMENT

Fix an integer $k\ge 0$. For a prime $p$, define \[ F_k(p)\ :=\ \prod_{i=0}^{k}(p^2+i). \] Let $P(n)$ denote the largest prime factor of $n$ (with $P(1):=1$).

Question. Is it true that for every fixed $k$ there exist infinitely many primes $p$ such that \[ P\left(F_k(p)\right) = p\ ? \]

Equivalently (for $p>k$ so that $p\nmid p^2+i$ for $1\le i\le k$), the condition is: \[ \max_{1\le i\le k} P(p^2+i)\ \le\ p. \] Indeed, the factor $p^2$ ensures that $p$ divides $F_k(p)$, and the requirement that no prime $>p$ divides any of the other factors forces $p$ to be the largest prime divisor.

2) QUICK LITERATURE/CONTEXT CHECK

The Erdős Problems forum thread for #383 lists the problem as open and notes that a positive answer would imply the second part of Problem #382.(https://www.erdosproblems.com/forum/thread/383 (accessed 2026-01-17).) It also gives a heuristic: the "probability" that an integer $n$ has no prime divisor $\ge n^{1/2}$ is $1-\log 2>0$, suggesting the answer should be yes.(Same source as above.) A comment in the thread mentions a conditional approach for the first nontrivial case using conjectures about Newman--Shanks--Williams primes.(Same source as above; comment by user "Dogmachine".)

3) ATTACK PLAN

  1. Re-express the condition as a simultaneous smoothness requirement: $p^2+i$ must be $p$-smooth for $1\le i\le k$.
  2. Produce computational evidence for small $k$ by searching primes $p$ up to various cutoffs.
  3. Heuristically, model the events "$p^2+i$ is $p$-smooth" as roughly independent with probability $\approx 1-\log 2$; then the expected count up to $X$ is $\asymp \frac{X}{\log X}(1-\log 2)^k$, which diverges for fixed $k$.
  4. The hard step is making any of this unconditional: it is a problem about smooth values of quadratic polynomials at prime arguments, well beyond current sieve technology in general.

4) WORK

Computational evidence.
For each fixed $k$, we searched primes $p$ up to a cutoff $P_{\max}$ and checked whether $P(p^2+i)\le p$ holds for all $1\le i\le k$. The table below records counts and the first few solutions found.
$k$search rangeprimes $p$ found (first few; count)
1$p\le 50000$$7,41,43,47,73,\dots$ (count $1265$)
2$p\le 50000$$41,157,211,313,421,\dots$ (count $311$)
3$p\le 50000$$443,599,1229,1277,1301,\dots$ (count $87$)
4$p\le 50000$$8663,15361,16223,23873,\dots$ (count $11$)
5$p\le 50000$$15361,43777,44531,45131$ (count $4$)
6$p\le 200000$$125441,143881$ (count $2$)
7$p\le 200000$none found (count $0$)

These computations support (but do not prove) the conjecture that for each fixed $k$ there may be infinitely many such primes $p$.

Heuristic.
Write $n=p^2+i\approx p^2$. The condition $P(n)\le p$ is essentially $P(n)\le n^{1/2}$. The Dickman--de Bruijn heuristic for smooth numbers suggests \[ \mathbb{P}(P(n)\le n^{1/2}) \approx \rho(2)=1-\log 2. \] Assuming (very optimistically) that the $k$ conditions for $i=1,\dots,k$ behave like independent random events of probability $1-\log 2$, one predicts that among primes $p\le X$ the expected number of solutions is \[ \sim \frac{X}{\log X}\,(1-\log 2)^k, \] which tends to infinity with $X$ for each fixed $k$.

5) VERIFICATION

  • The equivalence "$P(F_k(p))=p$ iff $P(p^2+i)\le p$ for all $1\le i\le k$" holds for all primes $p>k$; for $p\le k$ one must treat separately because $p$ can divide some $p^2+i$. The computational search handled the definition directly by checking prime factors of each $p^2+i$.
  • The computations above are finite and cannot distinguish "infinitely many" from "finitely many but many"; they serve only as evidence.

6) FINAL

UNRESOLVED.

  1. (i) Strongest partial result obtained. Reformulated the problem as requiring simultaneous $p$-smoothness of the $k$ consecutive integers $p^2+1,\dots,p^2+k$. Provided explicit computational evidence: solutions exist for $k=1,2,3,4,5$ with many primes below $50000$, and for $k=6$ at least two primes below $200000$.
  2. (ii) First hard gap. No known unconditional method currently proves that infinitely many primes $p$ make even a single quadratic value $p^2+1$ have all prime factors $\le p$ (i.e. be $p$-smooth), let alone $k$ simultaneous values.
  3. (iii) Top 3 next moves.
    1. Attempt a lower-bound sieve for primes $p$ for which $p^2+i$ has no prime factor in $(p, p^{1+\varepsilon}]$ and then control larger prime factors via additional arguments (a difficult "parity problem" barrier is expected).
    2. Seek conditional proofs under strong hypotheses on prime values/smooth values of polynomials (e.g. variants of Dickson/Schinzel, or very strong distributional conjectures).
    3. Explore connections to known families where $p^2+i$ factors in a controlled way (e.g. algebraic identities or Pell-type parametrisations) that might force $p^2+i$ to be smooth for infinitely many $p$.
  4. (iv) Why it might be true. Heuristics based on smooth numbers (and the constant $1-\log 2$) suggest a positive density of $p^2+i$ should have all prime factors $\le p$, and for fixed $k$ the expected number of primes $p\le X$ satisfying all $k$ constraints should still grow like $\frac{X}{\log X}(1-\log 2)^k$.

7) COMPLETION ESTIMATE

Estimated probability of a full resolution from these notes alone: 15%.