Question #b6464

1 Answer
Jan 28, 2017

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.