Does $n^2+1\nmid n!$ hold for infinitely many $n\in\mathbb N$?

Question

How many positive integers $ n $ satisfy that $ n^2+1 \nmid n! $; are there infinitely many?

See also https://math.stackexchange.com/questions/1365712/prove-that-there-exist-infinitely-many-integers-n20151-mid-n?noredirect=1&lq=1 — Bart Michels, Jun 02 '20 at 17:21

score 13 · Accepted Answer · edited Apr 13 '17 at 12:20

13

OEIS entry A144255 states that H. Iwaniec, Almost primes represented by quadratic polynomials, Invent. math. $47$, $1978$, p. $171$–$188$, proves that there are infinitely many primes or semiprimes of the form $n^2+1$. If $n^2+1$ is a prime, then it doesn't divide $n!$. If it's a semiprime, one of its two prime factors is $\ge\lceil\sqrt{n^2+1}\rceil=n+1$, so again it doesn't divide $n!$. Thus there are indeed infinitely many $n$ such that $n^2+1\nmid n!$.

Whether there are infinitely many primes of the form $n^2+1$ appears to be an open problem.

edited Apr 13 '17 at 12:20

Community

1

answered Mar 16 '13 at 05:12

joriki

238,052

Thanks, this is the first time I hear the Semiprimes : ) – easymath3 Mar 16 '13 at 05:19

Does $n^2+1\nmid n!$ hold for infinitely many $n\in\mathbb N$?

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