Question

Question #b6464

Answer 1

See below.

Explanation:

if `p` is an odd integer, `p ge 3` then

`(p,(p^2-1)/2,(p^2+1)/2)` is a primitive Pythagorean triple.

You can verify easily that `p^2 + ((p^2 - 1)/2)^2=((p^2+1)/2)^2`

then making `a=(p^2-1)/2` and `b=(p^2+1)/2` it follows

`2(p+a+1)=2(p+(p^2-1)/2+1)=2p+p^2-1+2=(p+1)^2`

Now being a Pythagorean triple, exactly one of `p,a` is divisible by `3` but `p` is prime `> 3` then `a` is divisible by `3`. Also `(p^2-1)` is divisible by `4`. The multiplicity by `12` attached to `a` is left as an exercise.