Triangle, perimeter = area (part 2)?

Let $a, b, c$ denote the sides of triangle $\Delta ABC$ . Recall Heron's formula for the area of a triangle with $3$ given sides:

$A = \sqrt{s(s - a)(s - b)(s - c)}$

where $s = \displaystyle \frac{a + b + c}{2}$ .

Setting this equal to the triangle's perimeter, we have

$\sqrt{(a + b + c)(b + c - a)(a + c - b)(a + b - c)} = 4(a + b + c)$

Let $x = b + c - a$ , $y = a + c - b$ , and $z = a + b - c$ . We know that $x, y, z$ are positive integers because of the triangle inequality . Now we have

$16(x + y + z) = xyz$

We see that $x, y, z$ must all be even integers. Let $x = 2p$ , $y = 2q$ , $z = 2r$ :

$\displaystyle \begin{aligned} pqr & = 4(p + q + r) \\ \implies pqr - 4p & = 4(q + r) \\ \implies p & = \frac{4(q + r)}{qr - 4} \end{aligned}$

Without loss of generality assume $p \geq q \geq r$ . Then we have $2p \geq 2q \geq q + r \implies qr \leq 12$ .

Since $4 < qr \leq 12$ , we may test integral values of $q$ and $r$ with $q \geq r$ to find all such triangles. Finally, we see that there is only $5$ such triangles:

$(a, b, c) = (5, 12, 13), (6, 8, 10), (6, 25, 29), (7, 15, 20), (9, 10, 17)$

$\implies \displaystyle \sum_{i \ = \ 1} p_i = 30 + 24 + 60 + 42 + 36 = \boxed{192}$

I wrote code to do this
Herons formula is simplified at the bottom