Successors of perfect squares are not divisible by predecessors of double even numbers Is there a proof for:
successors of perfect squares are not divisible by ( not multiple of ) predecessors of double even numbers, i.e., $n^2+1$ not divisible by $4k-1$, or $3+4k$?

If $n^2+1$ is divisible by a positive integer of the form $4k+3$, then it is divisible by a prime $p$ of the form $4k+3$. But it is a standard result of number theory that if $p$ is a prime of that form, then $-1$ is not a quadratic residue of $p$.
For a group-theoretic proof, note that if $n^2\equiv -1\pmod{p}$, then $n$ has order $4$. But the order of any group element divides the order of the group, and $4$ does not divide $4k+2$.
