Does a right integral triangle exist, in that the hypotenuse is twice the value of one of the sides?

Question

Assume that the sides are x, y and z respectively, with z being the hypotenuse. I want to find a right triangle in which x, y and z are all integers, and z is either 2x or 2y.

Answer 1

The quick answer is yes, because the hypotenuse is twice the value of one side in a 30-60-90 triangle.  But here's the algebraic analysis:

 

x² + y² = z²

 

z = 2x

 

x² + y² = (2x)²

x² + y² = 4x²

y² = 3x²

y = (√3)x      (Don't worry about the -√3 solution since we can't have negative lengths)

 

Yes, when the y side = x√3