proof of Pythagorean triples

If $a,b$, and $c$ are positive integers such that

$$a^2+b^2=c^2$$(1)

then $(a,b,c)$ is a Pythagorean triple. If $a,b$, and $c$ arerelatively prime in pairs then $(a,b,c)$ is a primitivePythagorean triple. Clearly, if $k$ divides any two of $a,b$,and $c$ it divides all three. And if $a^2+b^2=c^2$ then$k^2{a}^2+k^2{b}^2=k^2{c}^2$. That is, for a positive integer $k$, if$(a,b,c)$ is a Pythagorean triple then so is $(ka,kb,kc)$.Hence, to find all Pythagorean triples, it’s sufficient to findall primitive Pythagorean triples.

Let $a,b$, and $c$ be relatively prime positive integers suchthat $a^2+b^2=c^2$. Set

$$\frac{m}{n}=\frac{a+c}{b}$$

reduced tolowest terms, That is, $\gcd (m,n)=1$. From the triangle inequality$m>n$. Then

$$\frac{m}{n}b-a=c.$$(2)

Squaring both sides of (2) and multiplying through by $n^2$we get

$$m^2{b}^2-2mnab+n^2{a}^2=n^2{a}^2+n^2{b}^2;$$

which, after cancelling and rearranging terms, becomes

$$b\left(m^2-n^2\right)=a(2mn).$$(3)

There are two cases, either $m$ and $n$ are of opposite parity, orthey or both odd. Since $\gcd (m,n)=1$, they can not both beeven.

Case 1. $m$ and $n$ of opposite parity, i.e., $m\not\equiv \pm n(mod\mathrm{\hspace{0.17em}\hspace{0.17em}2})$. So 2 divides b since$m^2-n^2$ is odd. From equation (2), $n$ divides $b$.Since $\gcd (m,n)=1$ then $\gcd (m,m^2-n^2)=1$, therefore $m$ alsodivides $b$. And since $\gcd (a,b)=1$, $b$ divides $2mn$. Therefore$b=2mn$. Then

$$a=m^2-n^2,b=2mn,\text{and from(}\text{<span class="ltx\_text ltx\_ref\_tag">2</span>}\text{)},c=\frac{m}{n}\mathrm{\hspace{0.17em}2}mn-(m^2-n^2)=m^2+n^2.$$(4)

Case 2. $m$ and $n$ both odd, i.e., $m\equiv \pm n(mod\mathrm{\hspace{0.17em}\hspace{0.17em}2})$. So 2 divides $m^2-n^2$. Then by the same processas in the first case we have

$$a=\frac{m^2-n^2}{2},b=mn,and\mathit{\hspace{1em}}c=\frac{m^2+n^2}{2}.$$(5)

The parametric equations in (4) and (5) appearto be different but they generate the same solutions. To showthis, let

$$u=\frac{m+n}{2}\text{ and }v=\frac{m-n}{2}.$$

Then $m=u+v$, and $n=u-v$. Substituting those values for $m$ and$n$ into (5) we get

$$a=2uv,b=u^2-v^2,\text{and}\mathit{\hspace{1em}}c=u^2+v^2$$(6)

where $u>v$, $gcd(u,v)=1$, and $u$ and $v$ are of oppositeparity. Therefore (6), with a and binterchanged, is identical to (4). Thus since$(m^2-n^2,2mn,m^2+n^2)$, as in (4), is aprimitive Pythagorean triple, we can say that $(a,b,c)$ is aprimitive pythagorean triple if and only if there existsrelatively prime, positive integers $m$ and $n$, $m>n$, such that$a=m^2-n^2,b=2mn,\text{ and }c=m^2+n^2$ .