Question

find all the positive integers n such that n^2+1 is divisible by n+1

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

Step-by-step explanation:

Answer 2

Answer:

Simplify n^2+1 to (n+1)(n-1) + 2 and then solve the problem

Step-by-step explanation:

n+1 | n^2 + 1

=> n+1 | n^2 -1^1 +2

=> n+1 | (n+1)(n-1) +2 ....  using property :- x^2 - y^2 = (x+y)(x-y)

=> n+1 | 2 ... n+1 | (n+1)(n-1). So we cancel it out

Therefore, n+1 | 2

All possible positive n is 1