How to prove that there are no integers a,b such that $b^2=4a+2$
This seems like a very simple prof but when i tried to work through it i keep on hitting walls. I tried to prove this by contradiction saying that suppose there exists an $a,b\in\mathbb{Z}:b^2=4a+2$
then i worked through the proof as different cases, when a and b is odd, a,b is even and a is odd and b is even but i cant seem to get the answer.
If $b$ is odd then so is $b^2$, which is absurd since $4a+2$ is even.
If $b$ is even, say $b=2k$ we get $4k^2=4a+2$ hence $4(k^2-a)=2$, hence $2$ is a multiple of $4$, which is absurd.