The equation $a^2 - 4b = 3$ has no integer solution

Question

Prove that: $$a^2 - 4b \neq 3$$ for all integers $a$ and $b$.

I'm not sure how to find a way to prove this statement. Some help would be appreciated.

Answer 1

Assume that it is true. So $a^2-4b=3$.

Case I :When $a$ is even. Then $a=2k$ where $k$ is an integer. Thus we have, $4(k^2-b)=3$. LHS is even and RHS is odd. Contradiction.

CaseII: When $a$ is odd. Then $a=2k+1$ where $k$ is an integer. Then, $(2k+1)^2-4b=3$, hence we have $2(k^2+k-b)=1$. Again LHS even, RHS odd. Contradiction.

Answer 2

Take $\mod4$

LHS $\equiv 0,1 \pmod 4$ but RHS $\equiv 3$

So no solutions