# Simplest Proof of Euclid's Formula for Primitive Triples?

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?

• It looks like you didn't prove these are all the triples, only that they are each a triple. – coffeemath Feb 22 at 14:55
• Why would there be a right-angled triangle with hypotenuse $\sqrt{c}$ and integer legs? – user10354138 Feb 22 at 14:58
• 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. – user10354138 Feb 22 at 15:02
• Your claim is really - if there exist an integer relation between the sides than this is the triplet structure. – Moti Feb 22 at 18:45
• 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 $c=m^2+n^2$ – Peter Feb 23 at 9:41