Is it true that for every $k$ there are infinitely many primes $p$ such that the largest prime divisor of$$\prod _{0\le i\le k}(p^2+i)$$is $p$?

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Let $P(n)$ denote the largest prime divisor of $n$. Your condition [ P\left(\prod_{i=0}^{k}(p^2+i)\right)=p ] is equivalent to [ P(p^2+i)\le p\quad\text{for every }i=1,2,\dots,k, ] since the $i=0$ factor is (p^2), so $p$ certainly divides the product and hence the largest prime divisor is (\ge p).

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.