For integers $r,s$ can $4s-1$ be a factor of $r^{2}+1$

Question

For integers $r,s$ can $4s-1$ be a factor of $r^{2}+1$?

I have found the question from here "Integer solutions for $x^2-y^3 = 23$."

Answer 1

Hint $\ $ Suppose $\,4s\!-\!1\mid r^2\!+1.\,$ $\,4s-1\,$ has a prime factor $\,p\equiv -1\pmod 4\,$ [else all are $\equiv 1\,$ so their product $\equiv 1,\,$ not $\, -1\,$]. Write $\, p = 4k+3,\,$ so $\, \color{#0a0}{p\!-\!1 = 2(2k\!+\!1)},\,$ so by little Fermat

$p\mid 4s\!-\!1\mid \color{#90f}{r^2+1}\,\Rightarrow\,{\rm mod}\ p\!:\,\ \color{#90f}{r^2\equiv -1}\ \Rightarrow\ \color{#c00}1\equiv r^{\color{#0a0}{p-1} }\equiv (\color{#90f}{r^2})^{\color{#0a0}{2k+1}}\equiv (\color{#90f}{-1})^{2k+1}\equiv \color{#c00}{-1}$

therefore $\ p\mid 2=\color{#c00}{1-(-1)},\ $ contra $\,p = 4k+3\,$ is odd.