# Queer Diophantine Equation

Have arrived at this equation in attempting to factorize a polynomial, and was wondering if there are any solutions in integers (preferably) or rationals (worst case) for it:

$$(r^2 + s^2)(t^2 + v^2) = (r^2t^2 + s^2v^2) + (rv + st)^2$$

Any help much appreciated.

Muchos Gracias mes amigos.

• I guess it doesn't have any non-zero solutions, (After expanding both sides of the equation) is that right? Okay, could I expand it to include Gaussian numbers, or would that make no difference? – Joebloggs Sep 7 '17 at 9:34
• $$(rv+st)^2-r^2v^2-s^2t^2=2rvst$$ what kind of question is that??? – individ Sep 7 '17 at 10:20

## 1 Answer

After expanding you find : $2rstv=0$.

Since the equation is symmetrical in all variables, let's try for instance $r=0$.

This leads to : $s^2(t^2+v^2)=s^2v^2+(st)^2$ which are obviously equal. The other substitutions give the same result.

So the solutions are the quadruplets $(r,s,t,v)$ where at least one of the coordinate is zero, making it an infinity of solutions (integer or not).

• Thanks for the help. I think you confirmed it for me. – Joebloggs Sep 7 '17 at 10:04