The tutor shows that yesterday’s formulas to generate Pythagorean triples are valid.

In yesterday’s post I showed a way to generate Pythagorean triples x, y, z from an odd number n:

Let’s make sure that, generated as above, x²+y²=z²

n^2 +((n^2 -1)/2)^2 = n^2 + (n^4-2n^2+1)/4

which leads to, after getting a common denominator,

4n^2/4 + (n^4-2n^2+1)/4 = (n^4+2n^2+1)/4

Note also that

((n^2+1)/2)^2 = (n^4+2n^2+1)/4

Therefore

n^2 +((n^2 -1)/2)^2 = ((n^2+1)/2)^2

so the generating formula for pythagorean triples is proven.

Source:

Dudley, Underwood. Elementary Number Theory. New York:
  W H Freeman and Company, 1978.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.