Find another sum of squares for $3^{12}-6^6+2^{12}$ I have a question about factorization of number $3^{12}-6^6+2^{12}$. By completing the square one can show that$$3^{12}-6^6+2^{12} = (3^6-2^6)^2+6^6 = 665^2+216^2$$    If we can find another representation of this number as sum of squares then may try Euler's factorization method. But how to find another sum of squares in the simplest way? Maybe other methods work?

We have
\begin{eqnarray*}
665^2+216^2 = 488881 = 37 \times 13213 = (1^2+6^2)(a^2+b^2)=(a+6b)^2+(6a-b)^2.
\end{eqnarray*}
Now solve $a+6b=665,6a-b=216$ gives $a=53,b=102$ now swap $a$ and $b$ to get another representation as the sum of two squares.

 $$ 420^2+559^2 =488881=665^2+216^2.$$

