We may consider formulating a proof by contradiction.
Assume $\gcd(n^2+1,n!+1)=p$, then $p|(n^2+1)$ and $p|(n!+1)$.
Hence, we can conclude $n<p<n^2+1$ .
Can we go on from here?
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6https://math.stackexchange.com/questions/4580215/do-we-have-sum-n-1-infty-frac-gcd-left1n-1n2-rightn-stackrel – Barackouda Jul 01 '23 at 06:57
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What is the source of this question? – Dietrich Burde Jul 01 '23 at 11:49
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I guess we can prove it by an easier way. – Itoz Darbien Jul 01 '23 at 13:03
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You can conclude that any such $p$ is a prime number. If $p=qr$ then it would also be true that $q,r>n$ whence $p>n^2+1$. That may be why you chose $p$ to represent the gcd, but you didn't explicitly point it out. – Keith Backman Jul 02 '23 at 19:22