Show that there are infinitely many basic Pythagorean triples (x,y,z) in which z-x=2

z = (x+2)
if z (and x) even then y is also even so not basic

so z and x both are odd
let z = (2m+1) and x = (2m-1)

(2m+1)^2 - (2m-1)^2 = y^2

or 8m = y^2

so m= y^2/8

y ^2 = 16n^2 shall be a solution as y^2 is divisible by 8 and perfect square so divisible by 16

so m = 2n^2 and there is no other solution

so x = (4n^2-1), y = 4n, z = (4n^2+1) are solutions

there are infinite solutions and lowest 5 are below

n =1 gives (3,4,5)
n= 2 gives (15,8, 17)
n= 3 gives ( 35, 12, 37)
n = 4 gives (63,16,65)
n =5 gives (99,20,101)