According to this answer, $\sqrt{n!}$ has no integer solutions for $n>1$ due to Bertrand's postulate. However $\sqrt{n!+1}$ has integer solutions at 4,5, and 7. Is there any simple explanation why this is the case?
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$X^{2n}=0$ has integer solutions, but $X^{2n}+1=0$ has not. So this can always happen when you add $1$ somewhere. – Dietrich Burde Apr 30 '20 at 09:06
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We don't know much about the latter problem; further solutions are conjectured not to exist, but maybe they do. The best simple explanation we can give is that, because $n!$ has a very specific kind of factorization, there's a reason it's difficult for it to also other very specific kinds, such as those of squares. By contrast, $n!+1$ doesn't have that kind of special factorization. But $n!+1=m^2$ is equivalent to $n!=(m-1)(m+1)$, a useful factorization that has very different opportunities compared with $n!=m^2$, as we can put different prime factors of $n!$ into each of $m\pm1$. (Well, they'll both be even.)
Also, bear in mind that whether a Diophantine equation has no solutions at all or a handful of small ones is sometimes almost a matter of luck. What I mean by that is this: plenty of results hold for "all sufficiently large" positive integers, so we can often preclude solutions beyond a certain point, but in the worst case scenario earlier values have to be checked one by one. If anyone ever proves what all the solutions of $n!+1=m^2$ are, the proof will involve some logic that precludes all but the smallest solutions, but the few we know (plus maybe some more?) will just happen to manage it. Either that, or it'll turn out infinitely many solutions exist. (Mind you, that would contradict the abc conjecture, which is only a conjecture, but a seriously entertained one.)
J.G.
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