I have been reading about Pythagorean triples from the wiki page link here.
It says that a pythagorean triple consists of 3 positive integer's $ a, b, c $ such that $ a^2 + b^2 = c^2 $.
Also if all the integers in a triple say $ a, b, c $ are relatively prime then the triplet is called Primitive Pythagorean triplet.
As I was reading more in this article it also described about generating triple using Euclid's formula.
the formula is as follows :
$ a = m^2 - n^2, b = 2mn, c = m^2 + n^2 $ where $ m > n > 0 $
for any 2 values of $m$ and $n$ the above formula will give a Pythagorean Triple.
To get a Primitive Pythagorean triple, $m$ and $n$ have to co-prime and not both odd.
I wanted to understand the proof of this formula.
I don't understand this part of the proof which is also given in the wiki page.
"As ${{\tfrac {m}{n}}}$ is fully reduced, m and n are coprime, and they cannot both be even. If they were both odd, the numerator of ${\displaystyle {\tfrac {m^{2}-n^{2}}{2mn}}}$ would be a multiple of 4 (because an odd square is congruent to 1 modulo 4), and the denominator 2mn would not be a multiple of 4. Since 4 would be the minimum possible even factor in the numerator and 2 would be the maximum possible even factor in the denominator, this would imply a to be even despite defining it as odd. Thus one of m and n is odd and the other is even, and the numerators of the two fractions with denominator 2mn are odd. Thus these fractions are fully reduced (an odd prime dividing this denominator divides one of m and n but not the other; thus it does not divide m2 ± n2). One may thus equate numerators with numerators and denominators with denominators, giving Euclid's formula"
It uses unique fractionization $\Rightarrow$ uniqueness of reduced fractions (with denominators $> 0),\,$ i.e.
$\qquad\qquad \begin{align}\gcd(\color{#c00}{c,b})=1\\ \gcd(j,k)= 1\end{align}$, $\ \ \dfrac{c}b = \dfrac{j}k\ \Rightarrow\ \begin{align} c&\,=\,j\\ b &\,=\, k\end{align},\ \ \ {\rm for}\ \ b,c,j,k\in \Bbb Z,\ b,k > 0$
Follow the link for a simple proof using Euclid's Lemma (hint: $\,j = ck/b\,\Rightarrow\,\color{#c00}{b\mid c}\,k\,\Rightarrow\,b\mid k)$
Remark $ $ A more conceptual way to derive this parametrization of Pythagorean triples is to employ arithmetic of Gaussian integers $\,\Bbb Z[i] = \{ a + b\,i\,: a,b\in\Bbb Z\}$. Like integers they enjoy (Euclidean) division (with smaller remainder) and this implies they too satisfy the analog of the Fundamental Theorem of Arithmetic = existence and uniqueness of factorization into primes (= irreducibles). This implies that coprime factors of a square must themselves be squares (up to unit factors $\,\pm1,\pm i)$
Thus if $\ z^2 = x^2 + y^2 = (x-y\,i) (x+ y\,i) $ and $\,x,y\,$ are coprime then one easily checks that $\,x-y\,i,\,x+y\,i\,$ are coprime, so being coprime factors of the square $\,z^2$ they must themselves be squares (up to a unit factor). Thus e.g. $\ x + y\ i\, =\, (m + n\ i)^2 =\ m^2 - n^2 + 2mn\, i,\,$ hence $\,x = m^2-n^2,\ y = 2mn\,$ (using the unit factor $1$; using the other unit factors $\, -1,\pm i\,$ merely changes signs or swaps $\,x,y\,$ values). Notice how very simple the solution is from this perspective.
This is a simple prototypical (arithmetical) example of the simplification that results from the transformation of nonlinear problems into linear problems by working in larger algebraic extension rings. See here for some further discussion of such.