Question

Write a Pythagorean triplet whose one member is

$\left(i\right)6$

$\left(ii\right)14$

$\left(iii\right)16$

$\left(iv\right)18$

Answer 1

Pythagorean Triplet is given by $(2m,m^2-1,m^2+1)$ where $m$ is a positive integer.

(i) $6$

Let, $2m=6\Rightarrow m=\frac{6}{2}=3$

$\Rightarrow m^2-1=3^2-1=9-1=8$

$\Rightarrow m^2+1=3^2+1=9+1=10$

Therefore, Pythagorean Triplet is $(6,8,10)$

(ii) $14$

Let, $2m=14\Rightarrow m=\frac{14}{2}=7$

$\Rightarrow m^2-1=7^2-1=49-1=48$

$\Rightarrow m^2+1=7^2+1=49+1=50$

Therefore, Pythagorean Triplet is $(14,48,50)$

(iii) $16$

Let, $2m=16\Rightarrow m=\frac{16}{2}=8$

$\Rightarrow m^2-1=8^2-1=64-1=63$

$\Rightarrow m^2+1=8^2+1=64+1=65$

Therefore, Pythagorean Triplet is $(16,63,65)$

(iv) $18$

Let, $2m=18\Rightarrow m=\frac{18}{2}=9$

$\Rightarrow m^2-1=9^2-1=81-1=80$

$\Rightarrow m^2-1=9^2+1=81+1=82$

Therefore, Pythagorean Triplet is $(18,80,82)$