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Let $n>1$ and $a_n=(n!)^2+1$, Show that there is an odd prime number $p>n$ such that $p \mid a_n$

my attempt: if $p>n$ then $p-1 \geq n$ and so $(p-1)!=(p-1)(p-2)...n!$, and i tried to use Wilson's theorem which says that $(p-1)!=-1$ $[p]$, but it seems to be a closed door.

Bill Dubuque
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Hidda Walid
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HINT: $2$ does not divide $(n!)^2+1$ [because it is an odd number]. Meanwhile neither does any other integer $k \in \{3,\ldots, n \}$ [because $(n!)^2+1 \pmod k$ $ = 1$ for each such $k$].

So $(n!)^2+1$ has no divisors $2$ through $n$, including no prime divisors $2$ through $n$. What can you conclude about then the prime divisors of $(n!)^2+1$?

Mike
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    i see it now, we know that the smallest divisor of any number $\geq$2 is a prime number, so we conclude that the first divisor >n is the prime we are looking for – Hidda Walid Apr 04 '22 at 01:09
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    @HiddaWalid it is also possible for $a_n$ itself to be prime; indeed, $a_1,a_2,a_3,a_4,a_5$ are primes $ 2, 5, 37, 577, 14401.$ We finally get composite for $a_6 = 518401 = 13 \cdot 39877 ,$ next $a_7= 25401601 = 101 \cdot 251501 $ Alright, $a_9$ is prime again. – Will Jagy Apr 04 '22 at 01:47
  • Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Apr 04 '22 at 01:56