Answer to Question #350944 in Combinatorics | Number Theory for WSN

Question

1. Find all positive integers n such that n²+ 1 is divisible by n+ 1. 

Answer 1

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, hence

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$\iff$ a|b.

Then n+1|n²+1 $\iff$ 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.

Answer: 1