Find integer solution for square equation 
Does $a^2 + b^2 = 6 c^2$ have any integer solution?

My thought:
(0,0,0) is obviously a solution but I don't think there are any others.
Because if I take equation of modulo 6, it gives $a^2 + b^2 = 0 \pmod 6$
a and b can only be 3 mod 6 or 0 mod 6. However, I am not sure what is the next step here. Any help is appreciated

Consider a $\pmod{3}$ argument.
Since $6c^2$ is a multiple of $3$, you must have that
$a^2 + b^2$ is a multiple of $3$.
However, for any integer $n$, either
$n^2 \equiv 0 \pmod{3}$ or $n^2 \equiv (+1) \pmod{3}$.
Therefore, $a$ and $b$ must each be a multiple of $3$.
Let $r$ denote the largest positive integer exponent such that $3^r$ divides $a$.
Let $s$ denote the largest positive integer exponent such that $3^s$ divides $b$.
Since the constraint $a^2 + b^2 = 6c^2$ is symmetrical around $a$ and $b$, you can assume, without loss of generality, that $r \leq s$.
Since $3^r$ divides $a$ and $3^r$ divides $b$, you must have that $3^{(2r)}$ divides $6c^2.$
This implies that $3^{(2r - 1)}$ divides $c^2$.
This implies that $3^r$ divides $c$.
Therefore, you can let

*

*$d = \frac{a}{3^r}$.

*$e = \frac{b}{3^r}$.

*$f = \frac{c}{3^r}$.

Then, $~\displaystyle d^2 + e^2 = 6f^2$ 
where $d$ is not a multiple of $3$.
This implies that $d^2 \equiv 1\pmod{3}.$
Then, you will have that $e^2 \equiv 0\pmod{3}$ or
$e^2 \equiv 1\pmod{3}$.
In either case, you will not have that $d^2 + e^2$ is a multiple of $3$.
This yields a contradiction, since $d^2 + e^2 = 6f^2$.
