Interesting question

[terse grove]

Pls help i cant solve

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[terse grove]

It means which n positive integer
n! | 1^n +2^n … n^n is correct

[forest edge]

First observation I have is that when n is prime, we can take mod n on both sides, n! mod n = 0, and by fermat's little theorem a^n = a mod n, giving is n(n+1)/2, which is 0 mod n, as long as n is not 2.

If n+1 is prime, n! = n mod n+1, and by fermat's little theorem a^n = 1 mod n+1, giving us n. So the I'd guess that it works for prime n and prime-1 n, but obviously not sure

Just tried some values, that is not right, seems like for almost all values its false, now imma switch to counting the number of powers of 2.

The exponent of 2 in the prime factorization of n! is O(n)

If n is odd, then 1^n + 2^n ... n^n mod 2 is 1+0+1...+0+1, there is exactly one more 1 than 0, so if n is odd, the sum is odd. This means that n! cannot divide 1^n + ... n^n for any odd n

Sorry, that was wrong, what matters is the total number of 1's not the amount relative to 0, the total number of 1's is ceil(n/2) so if that is odd, then n! cannot divide the sum

For every prime, there should be a similar argument

Actually no, you want to take mod 2^k for all k

This will give you a bound on the exponent of 2 in the prime factorization of 1^n + 2^n ... n^n

Then you lower bound the exponent of 2^k in the prime factorization n!, and that will be asymptomaticly larger than the first bound

This would show that it can't be true for all but finitely many n

1 and 3 are likely the only values