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.

Put $a=6k^2-2$, $b=6k$ and $c=6k^2+1$. Then
$$\left(6k^2-2\right)^2+(6k)^2=\left(6k^2+1\right)^2+3.$$

Behind these solutions is the observation that $$\left(\frac{x+y}{2}\right)^2-\left(\frac{x-y}{2}\right)^2=xy$$ We can re-write the given equation as $c^2-a^2=b^2-3$ and choose $b$ such that $b^2-3$ is the product of two odd numbers $xy$. This allows us to choose $a=\frac{x-y}{2}$ and $c=\frac{x+y}{2}$.

