Pythagorean Triplets

In general we can represent the lengths of the sides of a Pythagorean triangle as

$d(m^{2} - n^{2}), 2dmn, d(m^{2} + n^{2})$

for integer $d \ge 1$ and coprime positive integers $m,n$ such that $m \gt n$ with precisely one of $m,n$ being even. The perimeter $k$ of such a triangle will then be

$k = 2dm^{2} + 2dmn = 2dm(m + n)$

with the aforementioned conditions on $d.m.n.$ This then serves as the generating formula for the desired values of $k,$ as listed here .

Uniqueness is a bit harder to generalize, I think. I'll look at a couple of examples for guidance. With $(d,m,n) = (1,3,2)$ we have $k = 2*3*5 = 30.$ We can only have $dm = 3$ and $m + n = 5$ , for if:

• $dm = 1$ then $d = m = 1,$ but $m$ must be at least $2$ ;

• $dm = 5$ then $d = 1, m = 5,$ (as $m \ge 2$ ), which makes it impossible for $m + n = 3, n \ge 1$ ;

• $dm = 15$ then $m + n = 1,$ which is not possible under our conditions.

With $dm = 3, m \ge 2$ we must then have $d = 1, m = 3$ and thus $n = 2.$ Thus a perimeter of $k = 30$ corresponds to a unique Pythagorean triplet, namely $(d,m,n) = (1,3,2) \Longrightarrow (a,b,c) = (5,12,13).$

However, $k = 168$ can be generated by $(d_{1},m_{1},n_{1}) = (2,6,1),$ corresponding to $(a_{1},b_{1},c_{1}) = (24,70,74),$ and by $(d_{2},m_{2},n_{2}) = (3,4,3),$ corresponding to $(a_{2}, b_{2}, c_{2}) = (21,72,75).$ The coinciding values result because both $d_{1}m_{1} = d_{2}m_{2}$ and $m_{1} + n_{1} = m_{2} + n_{2}.$

This is an interesting problem, more analysis of which is provided here .