# Unique factorization and the Pythagorean equation.

Let's say I have a Diophantine equation in $x$ [and possibly other variables], and I work the problem out to the point where I've shown that there are integers $a,b,c,d$ such that \begin{align} x &= a^2 + b^2 \\ x^2 &= c^2 + d^2. \end{align} Clearly, one solution [up to permutation] would be $(c,d) = (a^2-b^2,2ab)$. And if $x$ is prime, then that's the only non-trivial solution [because Fermat's two-square theorem guarantees that $x$ has a unique representation as the sum of two squares].

But when $x$ is composite, it may be that $a^2+b^2 = u^2+v^2$ for one or more [non-trivial] pairs of integers $u,v$ not equal [in any permutation] to $a,b$. Which is almost certainly to say $(c,d) \neq (a^2-b^2,2ab)$.

So I conclude that some form/consequence of “unique factorization” comes into play here, and must be handled accordingly.

Am I correct?

• What do you mean when you say "if $x$ is prime, then that's the only solution"? $(c,d)=(x,0)$ is always a solution of $x^2=c^2+d^2$. Oct 8 '15 at 18:45
• if x is composite and a sum of two non-zero squares, there may still be only one way .E.g. x=45=36+9. Oct 8 '15 at 18:51
• @BarryCipra: only non-trivial solution. Thanks for the correction. Oct 8 '15 at 19:35
• @user254665: Corrected. Thanks. Oct 8 '15 at 19:40

As an example, consider $x = 65 = 8^2 + 1^2$. We have $x^2 = 4225 = 52^2 + 39^2$, where $52 \neq 2 \cdot 8 \cdot 1$. This confirms my conjecture that I cannot assume $(c,d)=(a^2-b^2,2ab)$ [in some permutation].