Solving Pytagorean Equation $x^2+y^2=z^2$ / حل معادله فیثاغورث

              (Persian) به نشانه تحسین علیرضا صالحی گل سفیدی که کتابش بارها به من روحیه داده است.

                 As a sign of admiration for Alireza Salehi Golsefidy whose book has inspired me many times.

                A Pytagorean_triple consist of three positive integers $\;a,\;b\;$ and $\;c,\;$ such that

$$a^2+b^2=c^2\;. \tag{P}$$

          This is how the Wikipedia article about them begins. Being dissatisfied with the way the subject is presented in various works, I will present the version that seemed to me the most correct and complete.

          The presentation from page 29, from this source :

COHEN Henri

Explicit Metods for Solving Diophantine Equations

f.e., f.l., 2006

seemed to me the best.

          

          If the triplet $(a,b,c)$ verifies the equation $(P)$, then the triplets $(\pm a,\pm b,\pm c),\; (b,a,c),\;(d\cdot a,d\cdot b,d \cdot c)\;-\;d\in \mathbb{Z}\;-$ also verifies $(P)$. Our luck is that, if the $\mathbf{gcd}(a,b)=1$, then $\mathbf{gcd}(a,b,c)=1$, and converselly. Thus, the general solution in integer numbers of the equation 

$$x^2+y^2=z^2 \tag{1}$$

is obtained from all the primitive solutions - that is, those for which $\mathbf{gcd}(x,y,z)=1$ - by multiplying by an arbitrary integer factor.

          On page 29, loc. cit. we read:

          "Proposition 5.1  The general solution to the equation (1) is given by the

          two disjoint parametrisations

$$(x,y,z)=(2st,s^2-t^2,\pm(s^2+t^2)) \tag{2}$$

     and

$$(x,y,z)=(s^2-t^2,2st,\pm(s^2+t^2)) \tag{3}$$

          where $s$ and $t$ are two coprime integers of opposite parity."[ and $s>0$, n.CiP]

     Remark CiP The addition of the right parentheses, made by us, is necessary because the pairs of parameter $(s,t)$ and $(-s,-t)$ give the same result. If we impose in the relations (2) and (3) as $s>t>0$, then all the solutions in non-zero natural numbers are obtained.

               We show in the table below some positive solutions obtained by formula (2); also, on a red background, solutions that violate the condition of parameters $s$ and $t$ to be of different parities (thus obtaining non-primitive solutions).

             In the proof of the Proposition 5.1, made also in the line of the cited work, we need the following

               LEMMA  Let $y$ and $z$ be two odd integers. Then

$$(y,z)=\left ( \frac{y+z}{2},\frac{y-z}{2}\right );$$

                   consequently      $(y,z)=1\;\Leftrightarrow\; \left (\frac{y+z}{2},\frac{y-z}{2}\right )=1.$

           (Here, $(u,v)$ denotes the greatest common divisor of the numbers $u$ and $v$.)

                Proof of Lemma  Let $d:=(y,z)\;$; from Bézout's Lemma there exist integers $u$ and $v$ such that $y \cdot u+z \cdot v=d$. From here

$$2d=2yu+2zv=(y+z)\cdot (u+v)+(y-z) \cdot (u-v),$$ so $d=\frac{y+z}{2}\cdot (u+v)+\frac{y-z}{2} \cdot (u-v)$, and again with Bézout's Lemma, we have

$$d_1:= \left (\frac{y+z}{2},\frac{y-z}{2} \right )\;\mid \;d. \tag{4}$$

     Now, from $d=(y,z)$ result $y=d \cdot u_1,\;z=d \cdot v_1$, with $(u_1,v_1)=1.$ Hence $\frac{y \pm z}{2}=d\cdot \frac{u_1 \pm v_1}{2}$, so, $u_1$ and $v_1$ being odd, $\;d\;\mid\;\frac{y \pm z}{2}\;$ so we must have

$$d\;\mid \; d_1. \tag{4'}.$$

Relations (4) and (4') show that $d=d_1$.

(end of proof Lemma)$\square$

                   Proof of Proposition 

                   We will look for solutions in positive integers. 

               The $x$ and $y$ cannot both be odd, because then $x^2+y^2$ has the form $4k+2$, which cannot be a perfect square. Then exactly one is odd and the other, say $x$, is even.

          We rewrite the equation (1) like this: $\left ( \frac{x}{2} \right )^2=\underset{q}{\underbrace{\frac{z-y}{2}}} \cdot \underset{p}{\underbrace{\frac{z+y}{2}}}.$ The $y$ and $z$ being 

coprime, according to the Lemma, $q=\frac{z-y}{2}$ and $p=\frac{z+y}{2}$ are also coprime. Then, in the last equation we must have both $\frac{z \pm y}{2}$ be squares. 

$$q=\frac{z-y}{2}=t^2\,,\,\;p=\frac{z+y}{2}=s^2 \tag{5} .$$

The simple observation that $y=p-q$ and $z=p+q$ shows that $p,\;q$ must be of different parities (otherwise $y$ and $z$ would be even); the same are $s,\;t$.

          From (5) and further from $(x/2)^2=p \cdot q$ results the formulas (2).

$\blacksquare$