Show that $(x,y,z)$ is a primitive Pythagorean triple then either $x$ or $y$ is divisible by $3$.

Question

Show that $(x,y,z)$ is a primitive Pythagorean triple then either $x$ or $y$ is divisible by $3$.

The solution is given by 13.1.2:

But how do you become the 2 congruences?

Answer 1

Primitive Pythagorean triples are generally of two forms:

Form A: $(4i)^2 +(4i^2-1)^2=(4i^2+1)^2; i ∈ N $

$a=4,8,12,...4i$

$b=3,15,35,..4i^2-1$

$c=5,17,37,...4i^2+1$

Form B: $(2i+1)^2+[2i(i+1)]^2=[2i(i+1)+1]^2$

$a=3,5,7,...2i+1$

$b=4,12,24,...2i(i+1)$

$c=5,13,25,...2i(i+1)+1$

By arguing on $i$, it is not difficult to see that for any $i ∈ N$ there is always a number in triple which is a multiple of $3$ or $5$ or$ 3 and 5$:

Form A:

if $i=3k$, then $a=3k$

if $i=3k ± 1$, then $b=4(3k ± 1)^2-1=3p$

if $i=3k ± 2$, then $b=4(3k ± 2)^2-1=3p$

Form B:

if $i=3k $, then $b=2(3k)(3k+1)=3p$

if $i=3k+1$, then $a=2(3k+1)+1=3p$

if $i=3k+2$, then $b=2(3k+2)(3k+3)=3p$

The same argument can be used for $5$.