Bound for the equation $a^2 = b^2 + 4n$ For a fixed natural number $n$, I want to study the integer solutions of $$a^2 = b^2 + 4n.$$ There is always at least one solution, namely $$(n+1)^2 = (n-1)^2 + 4n.$$ There can be others; for example, the solutions of $a^2 = b^2 + 24$ are $$(a,b)
 = (\pm 7, \pm 5), \; \; (\pm 5, \pm 1).$$
Conjecture: there are no solutions $(a,b)$ with $|a| > n+1$.
I thought of factoring $4n = a^2 - b^2 = (a-b)(a+b)$. So there should definitely not be solutions $|a| > 4n$ since at least one of the factors would be greater than $4n$. Can we get all the way down to $n+1$?

If $|a|=n+2$ then $b^2=(n+2)^2-4n$, which is strictly greater than $n^2$ but less than $(n+2)^2$, hence it must be $(n+1)^2$, so $n^2+4=n^2+2n+1$ so $n=3/2$(not possible)
Now if $|a|=n+3$ then  $(n+3)^2-4n$ is a perfect square, which is strictly greater than $(n+1)^2$ but less than $(n+3)^2$, hence it must be $(n+2)^2$, so $n^2+2n+9=n^2+4n+4$ so $n=5/2$(not possible)
Now if $|a|>n+3$
$a^2>a^2-4n>(a-1)^2$
Hence it is never a perfect square as it's lying between two consecutive perfect squares, hence your conjecture is true.
