Solutions to equation (or proof that there is no solution)

Observe that not all of $a,b,c$ can be even otherwise the left side is divisible by $4$, whereas the right side is not. This means, exactly two of $a,b,c$ are odd and one of them is even.

Suppose both $a$ and $b$ are odd (same analysis will hold if both $c$ and $b$ are odd) and $c$ is even. In which case $a^2 \equiv b^2 \equiv 1 \pmod{4}$ and $c \equiv 0 \pmod{4}$, thus the left side is $0 \bmod 4$, whereas the right side is $2 \bmod 4$, so not possible.

This means the only possibility left is when both $a$ and $c$ are odd and $b$ is even. Say $a=2k+1, b=2s, c=2t+1$, then we have

$$(k^2+k)-s^2+(t^2+t)=0$$ For this one possibility is $k=t=1$ and $s=2$. That gives $3,4,3$ as a solution. However there are more possibilities such as $k=t=8$ and $s=12$. That gives, $17,24,17$ as another solution.