<s>[INST] A system of quadratic Diophantine equations with four variables Is the following system has any positive integer solution $(x,y,u,v)$?
$$\begin{cases} x^2+y^2=u^2\\ x^2-y^2=v^2 \end{cases}$$
I can prove that any pair of these integers can be relatively prime, but I couldn't find any solution. Any hint?
Thanks in advance!
 [/INST] From solutions of the Pythagorean equation we have $x = 2ab, y = a^2 - b^2$, so we must have $$(a^2 - b^2 - 2ab)(a^2 - b^2 + 2ab) = v^2$$ we have $(a^2 - b^2 - 2ab, a^2 - b^2 + 2ab) = 1$, so $$a^2 - b^2 - 2ab = (a - b)^2 - 2b^2 = x^2$$ $$a^2 - b^2 + 2ab = (a + b)^2 - 2b^2 = y^2$$ your equation has solution if and only if the Bell equations, $x^2 - 2b^2 = m^2$, $y^2 + 2b^2 = n^2$ have solutions for $(m, n)$ such that $2|m+n$, you can easily find all solutions of this equations and I think you can find some solutions this way.
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