A finding (?) about some primitive Pythagorean triples

Question

I have just stumbled on the fact that the sum of the three absolute differences between each pair of a primitive Pythagorean triple [absolute values of (a-b), (b-c) and (c-a), where a,b,c constitute the triple ] add up to the square of the integer that is one less than the smallest of the integers in the triple. Example 1: starting with the {3,4,5} triple, 4-3=1, 5-4=1, and 5-3=2; sum of the positive differences =1+1+2 = 4 = 2^2. Example 2: {7,24,25} triple; 24-7 +25-24 + 25-7 = 17+1+18 = 36 = 6^2 Moreover, I find that this result applies only to primitive triples where the least of the three integers is odd. For primitive triples where the lowest integer is even (actually a multiple of 4), as in [8,15,17}, {12,35,37}, [16,63,65}, etc. the sum of the three positive differences turns out to be twice a square number. I have a strong feeling that this can be proven. That is my next step.

Answer 1

Let the triple have smallest term odd. Call that term $a$. Then there exist integers $s$ and $t$ of opposite parity such that $a=s^2-t^2$, $b=2st$, and $c=s^2+t^2$. Our sum of absolute values is $$(2st-s^2+t^2)+(s^2+t^2-2st)+(s^2+t^2-s^2+t^2).$$ This is $4t^2$, That is certainly a perfect square, but it is not necessarily the square of $1$ less than the smallest element in the triple.

For example, let $s=7$ and $t=4$. Then we get the triple $(33, 56,65)$. Our sum of absolute values of differences is $64$, which is not the square of $32$.

In the case where the even side is smallest, our sum of absolute values is $$(s^2-t^2-2st)+(s^2+t^2-2st)+(s^2+t^2-s^2+t^2).$$ This is $2s^2-4st+2t^2$, which is indeed twice a perfect square.