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.
 A: 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 $a^2-3=c^2-b^2$.
A: 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 $c=\frac{x+y}{2}$.

A: Although the formula I posted.   Integral solutions of $x^2+y^2+1=z^2$
Thought and decided to write as simple as possible.  In equation:
$$a^2+b^2=c^2+q$$
$q$ - is specified by the problem statement. Then choose one solution $a$ so that it was possible to factor this way.
$$a^2-q=t(t+2s)$$
So the solution of this equation is always there. And can be written as:
$$b=s$$
$$c=s+t$$
A: Note that the requirement is simply to demonstrate infinitely many solutions - we are not required to find all solutions.
Setting $c=b+1$, we see that $a^2 = c^2-b^2+3 = 2b+4$
Therefore for any even $a>2$,  we can choose $b=\frac{a^2-4}{2}$ and $c=b+1$. This gives infinitely many solutions as required.
