Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$. Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$.
My attempt:
$(a+1)(a-1)+(b+1)(b-1)=c^2+1$
This form didn't help so I thought of $\mod 3$, but that didn't help either. Please help. Thank you.

In order that an integer $n$ can be represented as $c^2-b^2$, it suffices that $n\not\equiv{2}\pmod{4}$. In such a case $n$ can be written as the product of two divisors with the same parity, $n=pq$, and we can take $c=\frac{p+q}{2},d=\frac{p-q}{2}$. So it is sufficient to prove that for an infinite number of integers $a$, $a^2-3\not\equiv 2\pmod{4}$. This happens every time $a$ is even.
For instance, take $a=10$. We can write $n=a^2-3=97$ as $1\cdot 97$, hence
$$ (a,b,c) = (10,48,49) $$
is a solution of $a^2-3=c^2-b^2$.
