1

Is there a standard method to find positive integral solutions of the equation $x^2+y^2= qz^2$, where $q$ is a positive integral constant? If yes, then would someone please show it to me? Actually I had come across a problem in which we had to find integral solutions of $a^2+b^2=2c^2$ which was done by changing it to $(a-2/2)^2+(b-2/2)^2= c^2$ and using the parametric form of Pythogorean triplets but what if we had other numbers such as 4 or 5 in place of q?

shsh23
  • 1,135
  • The left hand side looks like an ellipse, by swapping x and y you would get another ellipse, so I think you get a family of solutions. – Emil Dec 19 '18 at 07:26
  • 1
    https://en.wikipedia.org/wiki/Hasse_principle – metamorphy Dec 19 '18 at 08:04
  • If you can find one solution then the method of parametrizing Pythagorean triples (draw lines with rational slope thru $(1,0)$ and find the other point of intersection with the unit circle) generalizes to this case as well. You will use the ellipse $x^2+py^2=q$ instead. – Jyrki Lahtonen Dec 19 '18 at 13:37
  • ^non-trivial solution :-) OTOH congruence consideration forbid the existence of solutions for some choices. For example, it is easy to show that $x^2+y^2$ is divisible by $3$ only if both $x$ and $y$ are multiples of three. This quickly allows you to conclude that $x^2+y^2=3z^2$ has no non-trivial solutions. The same works for any prime $\ell\equiv3\pmod4$ in place of $3$. – Jyrki Lahtonen Dec 19 '18 at 13:46
  • 1
    @Jyrki the rational slope method does give all rational solutions; multiplying out it gives a Pythagorean triple type of formula. If you then say to divide out by the gcd of x,y,z you do get all primitive triples. More work: for such a parametrization, with coprime $u,v$ the set of gcd's of $x,y,z$ is finite. For each, a new integer parametrization may be produced.. compare answer and my comment at https://math.stackexchange.com/questions/3044517/how-can-we-solve-the-diophantine-equation/3044603#comment6276439_3044603 – Will Jagy Dec 19 '18 at 17:54
  • 1
    https://artofproblemsolving.com/community/c3046h1051335_pythagorean_triple – individ Dec 21 '18 at 05:50

0 Answers0