Find all $c\in\lbrace{0,1,2,3,...\rbrace}$ such that $n^2+cn$ is a perfect square for all $n\in\mathbb{N}$.

Question

I would like to determine all non-negative integers $c$ (i.e. $c\in\lbrace{0,1,2,3,...\rbrace}$) such that $n^2+cn$ is a perfect square for all $n\in\mathbb{N}$. How can I find all of them?

Best wishes

Answer 1

This is false unless $c = 0$, so that $n^2+cn = n^2$. For $c > 0$, choosing $n = c$ gives $n^2+cn = 2c^2$, which is never a perfect square.

Answer 2

$0$ is the only such number.

If not, let $c(\neq 0)$ be such a number and $p$ be a prime. Then, $p^2+cp=p(p+c)$ is a perfect square. Now, for this to be true, $p \mid (p+c) \implies p \mid c$. Since this is true for all $p$, $c$ is divisible by all primes, which evidently is not possible.