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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 (https://projecteuler.net/problem=9. 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.

Siddhartha Ghosh
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2 Answers2

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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 $5740000 + 99917631+100082369=708750+102514400+102516850$.

Sarvesh Ravichandran Iyer
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Let $a = r^2 - s^2$, $b = 2rs$ and $c = r^2 + s^2$ be a triplet whose sum is equal to that of another triplet $l = x^2 - y^2$, $m = 2xy$ and $n = x^2 + y^2$. Then $a+b+c = l+m+n$; which gives

$$ r(r+s) = x(x+y) $$

The above equation is nothing but representing a number as the product of two factors in two different ways.

Thus if you take any number that has four or more divisors which can be represented as above then you can construct two Pythagorean triplet which have the same sum by plugging the values $r, s, l$ and $m$ in the above formula for $a,b,c,l,m$ and $n$.

Nilotpal Sinha
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