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A right triangle with side lengths $a,b,c$ where $c$ is the hypotenuse and using integers $m, n $ where $ m > n$, we can find Euclid's Formula.

First, given a right triangle with an hypotenuse $\sqrt{c}$ we can set its side lengths to be m and n, giving $c = m^2 + n^2$. Next, this works for all primitives, which can be proven given the greatest side length is always an odd number telling us that when m and n are even and odd, they will at minimum work for all primitives.

Using this we now can find the integer values of a and b.

$(m^2 + n^2)^2 = c^2$

$m^4 + 2m^2n^2 + n^4 = c^2$

$(m - n)^2 + (2mn)^2 = c^2$

Which results in Euclid's formula:

$(2mn, m^2 - n^2, m^2 + n^2)$

Is this proof valid?

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    $\begingroup$ It looks like you didn't prove these are all the triples, only that they are each a triple. $\endgroup$ – coffeemath Feb 22 at 14:55
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    $\begingroup$ Why would there be a right-angled triangle with hypotenuse $\sqrt{c}$ and integer legs? $\endgroup$ – user10354138 Feb 22 at 14:58
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    $\begingroup$ You don't know $c$ is a sum of two squares yet. You only know $c^2$ is a sum of two squares. It is on you to prove that in all such cases $c$ is indeed a sum of two squares. $\endgroup$ – user10354138 Feb 22 at 15:02
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    $\begingroup$ Your claim is really - if there exist an integer relation between the sides than this is the triplet structure. $\endgroup$ – Moti Feb 22 at 18:45
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    $\begingroup$ Not every odd integer is the sum of two squares of integers. In other words, not for every odd $c$, there are integers $m,n$ with $c=m^2+n^2$ $\endgroup$ – Peter Feb 23 at 9:41

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