Number of integer solutions of $a^2+b^2=10c^2$ Find the number of integer solutions of the equation $a^2+b^2=10c^2$.
I can only get by inspection that $a=3m, b=m,c=m$ satisfies for any $m \in Z$.
Is there a formal logic to find all possible solutions? Any hint?
Also i tried taking $a=p^2-q^2$, $b=2pq$ and $10c^2=(p^2+q^2)^2$
which gives $$\frac{p^2+q^2}{c}=\sqrt{10}$$ which is invalid, since a rational can never be an irrational.

Solving the $C$-function of Euclid's formula
$\quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$ for $(k), \space $ we can find Pythagorean triples for any given $C$-values, if they exist, that are primitive, doubles, or square multiples of primitives. This will not find, for example $(9,12,15)\space$ or $(15,20,25),$ but it will find $(3,4,5),\space (6,8,10),\space (12,16,20),\space (27,36,45), \space$ etc. We begin with the following formula. Any
$m$-value that yields an integer $k$-value indicates a valid $(m,k)$ pair for generating a Pythagorean triple.
\begin{equation}
C=m^2+k^2\implies k=\sqrt{C-m^2}\\
\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}.$
Here is an example for $C=40\implies 10c=4$ where $c$ is the one shown in the OP equation.
$$C=40\implies \bigg\lfloor\frac{ 1+\sqrt{80-1}}{2}\bigg\rfloor=4 \le m \le \lfloor\sqrt{40-1}\rfloor=6\\
\land \quad m\in\{6\}\Rightarrow k\in\{2\}\\$$
$$F(6,2)=(32,24,40)\implies (32,24,10\times 4)$$
This method will not find all Pythagorean triples that match the criteria but it will find an infinite number of triples that do such as:
$$c=1\longrightarrow (8,6,10\times 1)\\
c=2\longrightarrow (12,16,10\times 2)\\
c=4\longrightarrow (32,24,10\times 4)\\
c=9\longrightarrow (72,54,10\times 9)\\
$$
Note that any multiple of a triple found also yields a valid triple so
$3\times (8,6,10)\longrightarrow (24,18,10\times3)$ and provides the "missing" $c=3$ triple in the list above. The combination of the two will find all Pythagorean triples where the $c$ in $10c$ is an integer except for the most unusual case like $(3,1,10\times 1)$ mentioned in another post.
