Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. What is its general solution formula? Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. For examples,
$\begin{array}{l}
{1^2}+ {3^2} + {4^2}  = {1^2} + {5^2}\\
{1^2} + {2^2} + {6^2} = {4^2} + {5^2}\\
{322^2} + {562^2} + {597^2} = {601^2} + {644^2}\\
{938^2} + {1063^2} + {4722^2} = {536^2} + {4901^2}\\
 \vdots 
\end{array}$
What is its general solution formula? How to get the general solution formula?

It is clear that there are infinitely many solutions (see my comment above). We know that the identity $$(a^2+b^2+c^2+d^2)^2=(a^2-b^2-c^2-d^2)^2+(2ab)^2+(2ac)^2+(2ad)^2$$ gives all the solutions of $x^2+y^2+z^2+w^2=u^2$.
Now we have $$x^2+y^2+z^2=m^2+n^2\iff x^2+y^2+z^2+(im)^2=n^2$$ because the identity is valid in the ring of the Gaussian integers, so for the parameters $a,b,c,id$ we get $$(a^2+b^2+c^2-d^2)^2=(a^2-b^2-c^2+d^2)^2+(2ab)^2+(2ac)^2-(2ad)^2$$ so we have
$$\boxed{(a^2+b^2+c^2-d^2)^2+(2ad)^2=(a^2-b^2-c^2+d^2)^2+(2ab)^2+(2ac)^2}$$ This new identity giving consequently all the solutions of the proposed equation.
