Question

Find all integers n > 1 such that 1" + 2n +...+(n − 1)” 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. Then gcd(n+1,2)=1 (is this clear?), this is:

gcd(n+1,n2+1)=1 so that n+1,n2+1 are relative primes.

If n is odd, then n+1 is even, so gcd(n+1,2)=2, thus gcd(n+1,n2+1)=2. Remember that: If a>0, then gcd(a,b)=a⟺a|b. Then n+1|n2+1⟺gcd(n+1,n2+1)=n+1, then n+1 must be equal to 2, then n=1 is the only positive integer that satisfies the property.