Solve the Diophantine equation $6\left(x^{2}+y^{2}\right)=z^{2}+t^{2}$.
Since $6\left(x^{2}+y^{2}\right)=z^{2}+t^{2}$, we have $3 \mid z^{2}+t^{2}$. A square is always congruent to 0 or 1 modulo 3, so if the sum of two squares is divisible by 3, each of these squares is also divisible by 3. Therefore, we have $3 \mid z^{2}$ and $3 \mid t^{2}$, which implies $3 \mid z$ and $3 \mid t$. By writing $z=3 z^{\prime}$ and $t=3 t^{\prime}$ and substituting into the original equation, we get: $2\left(x^{2}+y^{2}\right)=3\left(z^{\prime 2}+t^{\prime 2}\right)$. By a similar reasoning as above, $3 \mid x^{2}+y^{2}$, and we can write $x=3 x^{\prime}, y=3 y^{\prime}$. Substituting this result into the equation, we find $6\left(x^{\prime 2}+y^{\prime 2}\right)=z^{\prime 2}+t^{\prime 2}$, which is exactly the original equation but with strictly smaller variables $x^{\prime}<x, y^{\prime}<y, z^{\prime}<z$ and $t^{\prime}<t$. By infinite descent, the only solution is the quadruplet $(0,0,0,0)$.