Proof: if $a$ and $b$ are integers, then $a^2-4b-3\neq 0$. I was wondering if someone could take the time to look over this proof and make sure it is correct. I greatly appreciate the help.
Proposition: If $a$ and $b$ are integers, then $a^2-4b-3\neq 0$.
Proof: Assume $a,b\in\mathbb{Z}$ and, for contradiction's sake, $a^2-4b-3=0$. Solving for $a^2$, we find $a^2=4b+3$. Clearly, $a^2 \equiv 3($mod $4)$.
Now, we can factor 2 out of the left-hand side of $a^2=4b+3$ yielding $a^2=2(2b+1)+1$. Thus, by the definition of odd, $a^2$ is odd. Since $a^2$ is odd, $a$ must be odd. By the definition of odd, we can write $a=2c+1$ where $c\in\mathbb{Z}$.
Now we can substitute for $a$ in $a^2$ to find $a^2=(2c+1)^2=4c^2+4c+1$. Factoring 4 out from the first two terms, we discover $a^2=4(c^2+c)+1$. Clearly, $a^2\equiv 1($mod $4)$. Earlier, however, we found that $a^2 \equiv 3($mod $4)$. Since $a$ can not be congruent to both 1 and 3 modulo 4, we have a contradiction. Therefore, if $a,b\in\mathbb{Z}$, then $a^2-4b-4\neq0$.

Yes, your proof is correct! Also, when you discorver that a has a remainder of 3 when divided by 4, you can go straight to the fact that a is odd.
