### Theory:

Objective:

In this lesson, we are going to learn about the Pythagorean triplets.

Pythagorean triplets:

A triplet $$a, b, c$$ of three natural numbers $$a$$, $$b$$, and $$c$$ is called a Pythagorean triplets, if ${a}^{2}+{b}^{2}={c}^{2}$.

For example:

I) $$(3, 4, 5)$$

$\begin{array}{l}{3}^{2}+{4}^{2}={5}^{2}\\ \\ \mathit{That}\phantom{\rule{0.147em}{0ex}}\mathit{is}\phantom{\rule{0.147em}{0ex}}9+16=25={5}^{2}\end{array}$

II) $$(15, 8, 17)$$

$\begin{array}{l}{15}^{2}+{8}^{2}={17}^{2}\\ \\ \mathit{That}\phantom{\rule{0.147em}{0ex}}\mathit{is}\phantom{\rule{0.147em}{0ex}}225+64=289={17}^{2}\end{array}$

To find a Pythagorean triplet:

We can write the Pythagorean triplet using this formula $2a,\phantom{\rule{0.147em}{0ex}}{a}^{2}-1,\phantom{\rule{0.147em}{0ex}}{a}^{2}+1$.

Here $2a,\phantom{\rule{0.147em}{0ex}}{a}^{2}-1,\phantom{\rule{0.147em}{0ex}}{a}^{2}+1$ is Pythagorean triplet for any natural number $a\succ 1$.

Note: If $$a$$ and $$b$$ are relatively prime natural numbers such that $a\succ b$ and exactly one of them is even and other is odd, and the three numbers contain no common factors.

For example: Find a Pythagorean triplet whose one number is 14.

We know that $2a,\phantom{\rule{0.147em}{0ex}}{a}^{2}-1,\phantom{\rule{0.147em}{0ex}}{a}^{2}+1$ is a Pythagorean triplet.

Now equate the given number with $$2a$$ to find the value of $$a$$.

$\begin{array}{l}2a=14\\ \\ \mathit{Then}\phantom{\rule{0.147em}{0ex}}a\phantom{\rule{0.147em}{0ex}}=\frac{14}{2}=7\end{array}$

Now substitute the $$a$$ in the known formula $2a,\phantom{\rule{0.147em}{0ex}}{a}^{2}-1,\phantom{\rule{0.147em}{0ex}}{a}^{2}+1$, then we get,

$\begin{array}{l}i\right)\phantom{\rule{0.147em}{0ex}}2a=2×7=14\\ \\ \mathit{ii}\right)\phantom{\rule{0.147em}{0ex}}{a}^{2}-1=\left({7}^{2}\right)-1=48\\ \\ \mathit{iii}\right)\phantom{\rule{0.147em}{0ex}}{a}^{2}-1=\left({7}^{2}\right)+1=50\end{array}$

Therefore the Pythagorean triplets are $$(14, 48, 50)$$.