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This question already has an answer here:

Can a square number be split as the sum of two other squares in two different ways? The Pythagorean numbers help us to identify numbers which can be split as sum of two squares. Is this unique?

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marked as duplicate by Saad, Jyrki Lahtonen, Ben, Lord Shark the Unknown, apnorton Dec 17 '18 at 7:07

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By the Brahmagupta–Fibonacci identity, which states that $$(a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$ you can get a number which is a sum of two squares two different ways by multiplying two sums of two squares.

So from two Pythagorean triples $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ , the left hand side simplifies to $(a_1a_2)^2$, you can get a number that can be written as a sum of two squares.

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  • $\begingroup$ Love whipping out the Brahmagupta-Fibonacci identity. math.stackexchange.com/questions/1012733/… $\endgroup$ – Matt Samuel Dec 17 '18 at 2:40
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    $\begingroup$ Of course if you have $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ then you also have $(a_2a)^2 + (a_2b)^2 = (a_1a_2)^2$ and $(a_1c)^2 + (a_1d)^2 = (a_1a_2)^2.$ This is how $65^2$ comes to be the sum of two squares in four different ways: take $a=4,b=3,c=12,d=5.$ $\endgroup$ – David K Dec 17 '18 at 12:00
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HINT

$3,4,5$ is a PPT, meaning that any multiple (like $6,8,10$) is a PT

$5,12,13$ is a PPT as well ...

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    $\begingroup$ Why the downvote? Not a good hint? $\endgroup$ – Bram28 Dec 17 '18 at 2:02
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For example \begin{eqnarray*} 65^2=63^2+16^2=33^2+56^2. \end{eqnarray*}

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  • $\begingroup$ In fact this happens infinitely often, see this sequence of the Online Encyclopedia of Integer Sequences. $\endgroup$ – Brian Hopkins Dec 17 '18 at 2:26
  • $\begingroup$ In addition, there are the obvious $65^2=26^2+52^2=25^2+60^2$ (multiples of $3,4,5$ and $5,12,13$). So that's not just two, but four ways for $65^2.$ $\endgroup$ – David K Dec 17 '18 at 2:43
  • $\begingroup$ @BrianHopkins Most of the numbers in that list are not squares, but of course some are. $\endgroup$ – David K Dec 17 '18 at 2:46
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    $\begingroup$ @DavidK Oops, yes: of the many related sequences, I think A009177 is more relevant. $\endgroup$ – Brian Hopkins Dec 17 '18 at 3:15
  • $\begingroup$ Thanks for the solution and illuminating discussion $\endgroup$ – Narayanan Raman Dec 30 '18 at 1:03

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