Pythagorean Triplets

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We seek to identify all positive integers a, b, and c so that a²+b²=c². A triple (a,b,c) satisfying this equation is called a pythagorean triplet. Since any multiple of a pythagorean triplet is again one, we only seek primitive ones where a, b, and c have no common factor. By setting $x=\frac{a}{c}$ and $y=\frac{b}{c}$, we can first seek rational numbers x and y so that x²+y²=1. =2in

Now, if (x,y) is such a pair of rational numbers, the line from (x,y) to (-1,0) will have rational slope - call it t. Along this line, y=t(x+1), so x² + t²(x+1)²=1 telling us that

(1+t²)x² + 2t² x = 1-t²

or

$$x^2 + \frac{2t^2}{1+t^2} x = \frac{1-t^2}{1+t^2}$$

Complete the square on the above equation giving

$$\begin{eqnarray*} (x + \frac{t^2}{1+t^2})^2 &= & \frac{1-t^2}{1+t^2}+\frac{t^4}{(1+t^2)^2}\\ & = & \frac{1}{(1+t^2)^2}\\ \end{eqnarray*}$$

Taking square roots gives

$$x=\frac{1-t^2}{1+t^2}.$$

Write $t=\frac{u}{v}$ with u, v positive integers having no common factor. This gives

$$\begin{eqnarray*} x &= & \frac{v^2-u^2}{v^2+u^2}\\ y &= & \frac{2uv}{v^2+u^2}\\ \end{eqnarray*}$$

Using $x=\frac{a}{c}$ and $y=\frac{b}{c}$ we see that

$$\begin{eqnarray*} a &= &k(v^2-u^2) \\ b &= &k(2uv) \\ c &= &k(v^2+u^2) \\ \end{eqnarray*}$$

for some rational number k. By thinking about the highest power p^r of an odd prime appearing in the denominator of k (written in lowest terms), one sees that p^r divides uv, u²+v², -u²+v², 2u², and 2v². This means $p^{\frac{r}{2}}$ divides both u and v contradicting u and v having no common factor, UNLESS r=0. Similarly by noting exactly one of a and b can be odd, and choosing b to be the even number, we can eliminate powers of 2 in the denominator of k. This shows that k is a whole number and we obtain: Theorem: The pythagorean triplets with b even are exactly the triples of the form

$$\begin{eqnarray*} a & =& v^2-u^2\\ b&=&2uv \\ c & =& v^2+u^2\\ \end{eqnarray*}$$

where u and v are positive whole numbers with no common factor.