Answer to Question #134202 in Discrete Mathematics for Promise Omiponle

Question

Prove that there are infinitely many solutions in positive integers x, y, and z to the equation x^2+y^2=z^2.
Hint: Let x=m^2-n^2 ,y= 2mn, and z = m^2+n^2, where m and n are integers.

Answer 1

If $x=|m^2-n^2|,\ y=2mn$, then

$x^2+y^2=(m^2-n^2)^2+(2mn)^2=\\ (m^2)^2-2m^2n^2+(n^2)^2+4m^2n^2=\\ m^4-2m^2n^2+n^4+4m^2n^2=\\ m^4+2m^2n^2+n^4=(m^2+n^2)^2=z^2$ .

Since there are infinitely many pairs $(m=1,\ n=i),\ i=1,2,....$ and

$y_i=2\cdot1\cdot i\neq y_j=2\cdot1\cdot j,\\ z_i=1^2+i^2\neq z_j=1^2+j^2\ for\ i\neq j$, then the equation has infinitely many integer solutions.