I know that we are gonna need to use one of the identities that the $\gcd$ is equal to but I can't remember one that would be useful for this problem. Any help?
2026-04-07 05:55:40.1775541340
Prove that for every natural number $n$ we have $\gcd(n! + 1, (n + 1)! + 1) = 1$
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Just note that a common divisor $d$ of $(n+1)!+1$ and $n!+1$ divides also $$ (n+1) (n!+1) - ((n+1)!+1) = (n+1)! + n+1 - (n+1)! - 1 = n. $$ And then $\gcd(n!+1, n ) = 1$, as $n!+1$ divided by $n$ leaves remainder $1$.