Infinitely many $ n \in \mathbb{N} $ such that $ n^2+1 $ has two divisors $ a,b $ such that $a-b=n $ Prove that there is infinitely many $ n \in \mathbb{N} $ such that $ n^2+1 $ has two divisors $ a,b $ such that $a-b=n $.
It is obvious that if $ p\mid n^2+1 $ then $\gcd(p,n)=1$. I tried to use the Chinese remainder theorem, but I got nothing. Please help me.

It works for $b,a$ consecutive terms in the sequence $x_j$ beginning
$$ 1, 2, 5, 13, 34, ..$$
with
$$ x_{j+2} = 3 x_{j+1} - x_j $$
which are (every second) Fibonacci numbers.
Although the problem does not require this, the product is precisely the given number $n^2 + 1.$  Thus the hint that $a^2 - 3ab + b^2 = -1$ really did finish the matter.
