Find all solutions to the equation x^2 + 2y^2 = w^2 with x > 0, y > 0, w > 0?

Question

A solution that relates to Pythagorean triples is preferred.

Answer 1

no one can if x y and w are real numbers there is infinite number of answers

Answer 2

x increases by increment of 3 each time starting from 3

y increases by increment of 4 each time starting from 4

z increase by increment of 5 each time starting from 5

and the starting set of 3,4,5 is a pythagorean triple so each triple is going to be a multiple of 3 for x, 4 for y and 5 for z

x = 3 + 3u

y= 4+ 4u

z= 5+ 5u

for each integer where u>0 there is a solution set of (x,y,z)

hope this helps