Question

If ${}^{\prime }{n}^{\prime }$ is odd no & $n>1$, them prove that $(n,\frac{n^2-1}{2},\frac{n^2+1}{2})$ is a Pythagorean, triplet. Write two Pythagorean triplet taking suitable value of ${}^{\prime }{n}^{\prime }$.

Answer 1

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

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

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

$\Rightarrow$ These form Pythagoras triplet.

For e.g., for $n=2$, & $n=3$

${\left(\frac{3}{2}\right)}^2+2^2={\left(\frac{5}{2}\right)}^2$ $4^2+3^2=5^2$

$(\frac{3}{2},2,\frac{5}{2})$ $(4,3,5)$.