I’m looking for a complete solution (parameterization or other) to the equation in the title, i.e., $$ U^2+V^2=A^2+sB^2, $$ where $s$ is squarefree [if necessary]. When $s=1$, the solution is well-known (and easy to derive), so we can assume $s \ne 1$.

Any references would be appreciated.

  • $\begingroup$ I don't seem to know the well known solution for $s=1.$ $\endgroup$ – Will Jagy Mar 15 '17 at 19:16
  • 2
    $\begingroup$ @WillJagy: If $u,v,a,b$ are integers, with $uvab \ne 0$, such that $u^2+v^2 = a^2+b^2,$ then there exist integers $g,h,m,n,t$ such that \begin{align} u &= \tfrac{1}{2}t(hn+gm), &\qquad\qquad&& a &= \tfrac{1}{2}t(hn-gm), \\ v &= \tfrac{1}{2}t(hm-gn), &\qquad\qquad&& b &= \tfrac{1}{2}t(hm+gn), \end{align} where $t$ is the greatest common divisor of $u,v,a,b$, and $\gcd(m,n)=1$. $\endgroup$ – Kieren MacMillan Mar 15 '17 at 19:19
  • $\begingroup$ can't see anything; will try refreshing the screen in a minute. Dickson's History, Volume 2, page 254, says Welsch (1910) claimed the full solution is... alright, same as what you type. $\endgroup$ – Will Jagy Mar 15 '17 at 19:21
  • $\begingroup$ On page 503, Fricke and Klein suggest $(z_1z_4 - z_2 z_3)$ as a normal form for an isotropic quaternary quadratic form, signature $++--.$ Seems to give Welsh's result without trouble... $\endgroup$ – Will Jagy Mar 15 '17 at 20:01
  • $\begingroup$ hmmm. appears that helps for $U^2 + s V^2 = A^2 + s B^2.$ Your version, not so much. You might try $u^2 + v^2 = a^2 + 27 b^2,$ where things about the cube root of two should appear. Or, worse, $u^2 + v^2 = a^2 + 23 b^2,$ $\endgroup$ – Will Jagy Mar 15 '17 at 20:28

For the equation.


You can write such a parameterization.





  • $\begingroup$ Is it exhaustive though? $\endgroup$ – MathGod May 22 '17 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.