Question

Find all integers n>1 such that 1^n +2^n+...+(n-1)^n is divisible by n​

Answer 1

Answer:

gcd(n+1,n2+1)=gcd(n+1,n2+1−n(n+1))=gcd(n+1,1−n)=gcd(n+1,1−n+(1+n))=gcd(n+1,2). If n is even, then n+1 is odd.

Step-by-step explanation:

extra questions to help you

Answer 2

Given:

$1^n+2^n+...(n-1)^n$

Firstly, $p=p^-^n$ mod $n$

For odd number, $n > 1$

$A:\sum_{p=1}^{n-1}p^n\equiv \sum_{p=1}^{n}p^n\equiv \sum_{p=1}^{n}(p-n)^n\equiv -\sum_{p=1}^{n}(n-p)^n$

$=-\sum_{p=1}^{n-1}p^n=-p$

$A\equiv -p$ mod $n$

$=p2A\equiv 0$ mod $n$

Therefore, either $2\equiv 0$ mod $n$ (which is impossible)

or, $A\equiv 0$ mod $(n)$  [which is correct]

So sum is divisible by odd $n.$

But for 0  even we will not get $A\equiv 0$ mod $(n)$

So correct answer is $(3,5,7,...)$