Question

What is the value of n for which 8, $(n^2-1)$, 17 are Pythagorean triplet.

• A. 16
• B. 15
• C. 4
• D. 8

Answer 1

The correct option is C

4

By Pythogoras theorem,
${17}^2=8^2+\left(\right(n^2-1){)}^2$
$\Rightarrow \left(\right(n^2-1){)}^2=225$
$\Rightarrow (n^2-1)=15$
$\Rightarrow n^2=16$
$\Rightarrow n=4$