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 $c$ are relatively prime in pairs then $(a,b,c)$ is a primitive Pythagorean 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 find all primitive Pythagorean triples.

Let $a,b$, and $c$ be relatively prime positive integers such that $a^{2}+b^{2}=c^{2}$. Set

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

reduced to lowest 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, or they or both odd. Since $\gcd(m,n)=1$, they can not both be even.

Case 1. $m$ and $n$ of opposite parity, i.e., $m\not\equiv\pm n(\,mod\,\,2\,)$. 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 $m$ also divides $b$. And since $\gcd(a,b)=1$, $b$ divides $2mn$. Therefore $b=2mn$. Then

 $a=m^{2}-n^{2},\quad b=2mn,\quad\mbox{and from (\ref{eq:p2})},\,\,c=\frac{m}{n}\,2mn-(m^{2}-n^{2})=m^{2}+n^{2}.$ (4)

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

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

The parametric equations in (4) and (5) appear to be different but they generate the same solutions. To show this, let

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

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

 $a=2uv,\,\,\,b=u^{2}-v^{2},\quad\mbox{and}\quad c=u^{2}+v^{2}$ (6)

where $u>v$, $gcd(u,v)=1$, and $u$ and $v$ are of opposite parity. Therefore (6), with a and b interchanged, is identical to (4). Thus since $\left(m^{2}-n^{2},2mn,m^{2}+n^{2}\right)$, as in (4), is a primitive Pythagorean triple, we can say that $(a,b,c)$ is a primitive pythagorean triple if and only if there exists relatively prime, positive integers $m$ and $n$, $m>n$, such that $a=m^{2}-n^{2},\,\,b=2mn,\,\,\mbox{ and }\,\,\,c=m^{2}+n^{2}$ .

Title proof of Pythagorean triples ProofOfPythagoreanTriples 2013-03-22 14:28:05 2013-03-22 14:28:05 fredlb (5992) fredlb (5992) 9 fredlb (5992) Proof msc 11-00