# Question #b6464

Jan 28, 2017

See below.

#### Explanation:

if $p$ is an odd integer, $p \ge 3$ then

$\left(p , \frac{{p}^{2} - 1}{2} , \frac{{p}^{2} + 1}{2}\right)$ is a primitive Pythagorean triple.

You can verify easily that ${p}^{2} + {\left(\frac{{p}^{2} - 1}{2}\right)}^{2} = {\left(\frac{{p}^{2} + 1}{2}\right)}^{2}$

then making $a = \frac{{p}^{2} - 1}{2}$ and $b = \frac{{p}^{2} + 1}{2}$ it follows

$2 \left(p + a + 1\right) = 2 \left(p + \frac{{p}^{2} - 1}{2} + 1\right) = 2 p + {p}^{2} - 1 + 2 = {\left(p + 1\right)}^{2}$

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 $\left({p}^{2} - 1\right)$ is divisible by $4$. The multiplicity by $12$ attached to $a$ is left as an exercise.