Question

In $n$ is an odd number and n > 1, then prove that $(n,\frac{n^2-1}{2},\frac{n^2+1}{2})$ is a Pythagorean triplet. Write two Pythagorean triplet making suitable value of $n$.

Answer 1

$n,\frac{n^2-1}{2},\frac{n^2+1}{2}$ are three numbers

$n^2+{\left(\frac{n^2-1}{2}\right)}^2$$=n^2+\frac{n^4-2n^2+1}{4}$

$=\frac{4n^2+n^4-2n^2+1}{4}$

$=\frac{n^4+2n^2+1}{4}$

$=\frac{(n^2{)}^2+2(n^2\left)\right(1)+(1{)}^2}{4}$

$=\frac{(n^2+1{)}^2}{(2{)}^2}={\left(\frac{n^2+1}{2}\right)}^2$

$\therefore n^2+{\left(\frac{n^2-1}{2}\right)}^2={\left(\frac{n^2+1}{2}\right)}^2$

It is a pythagoras triplet as the sum of square of two number is equal to the square of three numbers.

let $n=5$

$LHS=(5{)}^2+{\left(\frac{25-1}{2}\right)}^2=25+144=169=(13{)}^2$

Also, $RHS={\left(\frac{5^2+1}{2}\right)}^2={13}^2$

$\therefore 5,12,13$

let $n=7$ similarly we get,

$\therefore 7,24,25$