Question

Find total no. of integers n such that n2+1 is divisible by n+1

Answer 1

Answer :

n = 1

Step-by-step explanation:

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.

Answer 2

Solution:

First note that n2 – 1 = (n + 1)(n – 1).

So, (n + 1)|n2 – 1 for all integer n except (–1).

Now, suppose (n+1)| (n2 + 1).

Therefore, n+1 | (n2 + 1) – (n2 – 1)=2

⇒ n+1 is a factor of 2.

So, possible values (n+1) can be 1, 2, (–1) or (–2).

Therefore, possible values of n are 0, 1, (–2) or (–3).

Verify that each of them satisfies the given condition.