Pythagorean triples with same sum Can there be more than one Pythagorean triple such that the triple sum (a+b+c) is same? If yes, please provide some examples. Context: While solving Problem 9 of Project Euler (< This asks to provide the triplet product for a particular triplet sum 1000.), tried to write a generic program, which can give such product(s) for any triplet sum. Though for sum = 1000, there is exactly one such triplet, was not sure if this is the case for any triplet sum. Hence the question.

Given $x,y,a,b$ such that $x^2 + xy = a^2+ab$, with $x > y$ and $a>b$.

$2(x^2+xy) = 2(a^2+ab) \implies (x^2+y^2) + 2xy + (x^2-y^2) = (a^2+b^2) + 2ab + (a^2-b^2)$. The three terms on each side form a triple.

For example:

Let $x=8,y=7,a=10,b=2$. Then, $113+112 + 15 = 104+40+96$. Furthermore, $15^2 + 112^2 = 113^2$ and $40^2+96^2=104^2$.

More exciting: Let $x=48,y=44,a=64,b=5$. Then, $4224+ 368 + 4240 = 640+4071+4121$. Further $4224^2 + 368^2 = 4240^2$ and $640^2+4071^2=4121^2$.

Even bigger: Let $x=87,y=43,a=78,b=67$. Then, $7482+ 5720 + 9418 = 10452+1595+10573$. Further $7482^2 + 5720^2 = 9418^2$ and $10452^2+1595^2=10573^2$.

Finally, the biggest: $x=99,y=61,a=96,b=69$. Then, $12078+ 6080 + 13522 = 13248+4455+13977$. Further $12078^2 + 6080^2 = 13522^2$ and $13248^2+4455^2=13977^2$.

You can explore further.

EDIT : Just adding another : $x=10000 ,y= 287 ,a=10125 ,b= 35$ , with $5740000 + 99917631+100082369=708750+102514400+102516850$.