Is it possible to show that :for $n \geq 1:\sigma(n!-1) $ never be prime and why $\sigma(n!-1)\bmod 10 $ at most is $0$?

Question

This question is related to my question here , I w'd like to check if $n \geq 1:\sigma(n!-1) $ never be prime according to some computations which i did in wolfram alpha to come up with parity of sum power divisor function at $n!-1$ for some integer $n$ i observed that $\sigma(n!-1)\bmod 10 $ at most is $0$ mayeb give us somethings to answer the question " when is $\sigma(n!-1) $ perfect square .

Now my question here is:

How do i show that :$n \geq 1:\sigma(n!-1) $ never be prime and why $\sigma(n!-1)\bmod 10 $ at most is $0$ ?.

Note: The motivation of this question is to confirme if the last digits $\sigma(n!-1) $ is always $0$ for large $n$.

Thank you for any help

Answer 1

If $n\ge 4$, then $24 \mid n!$. It is an easy exercise to show that if $24 \mid N$, then $24 \mid \sigma(N-1)$. (Pair each factor of $N-1$ with its cofactor, and use that every unit modulo $24$ is its own inverse.) See, e.g.,

https://math.stackexchange.com/questions/1492606/suppose-that-n-is-an-integer-divisible-by-24-show-that-the-sum-of-all-the-posit

Answer 2

Suppose $\sigma(n!-1)$ is prime. Then, because $\sigma$ is a multiplicative function, $n!-1$ is a prime power $p^a$. Indeed, as Jan-Christoph Schlage-Puchta argues in his answer, $a\geq 3$. Since $p^a+1=n!$, the abc conjecture tells us that for any $\epsilon>0$ and all but finitely many possible counterexamples, the following inequality holds: $$p^a<n!<\operatorname{rad}(p^an!)^{1+\epsilon}=p\operatorname{rad}(n!)^{1+\epsilon}\\ p^{a-1-\epsilon}<\operatorname{rad}(n!)^{1+\epsilon}\\ \operatorname{rad}(n!)>p^{a/(1+\epsilon)-1}.$$ We can pick $\epsilon$, such that for all $a\geq 3$, we have $\frac{a}{1+\epsilon}-1>\frac{a}{2}+\epsilon$, so that $$\operatorname{rad}(n!)>p^{a/2+\epsilon}>\sqrt{p^a+1}=\sqrt{n!}.$$

So if we prove the above inequality is false for large $n$, we will get what we want. But we have $\operatorname{rad}(n!)=n\#$ (primorial), which is less than $4^n$ (this is a well-known inequality, see e.g. here), while $n!$ is asymptotically $\frac{n^n}{e^n}$ (Stirling's approximation), which quickly outgrows $(4^n)^2$, so the inequality cannot hold.

Hence, under the assumption of the abc conjecture, there are only finitely many $n$ such that $\sigma(n!-1)$ is prime. I can imagine one would expect there to be none, but I think we are far from an unconditional proof.

Answer 3

Probability tells you that there should be no solution, unless $n$ is very small. Since $\sigma$ is multiplicative, $\sigma(n!-1)$ can only be prime, if $n!-1$ is a prime power. Since $n!$ is not a prime, the case that $n!-1$ is prime can also be excluded. Since $n!-1\equiv 3\pmod{4}$, a prime square is also impossible. We conclude that a necessary condition for the statement "$\sigma(n!-1)$ is prime" is "$n!-1=p^a$ with $p$ prime and $a\geq 3$". Since cubes of primes are rare, no solution should exist. However, proving this statement is either easy, because some factorization or congruence argument works, or almost impossible.