Question

Write a Pythagorean triplet whose one member is.
i) $6$
ii) $14$
iii) $16$

Answer 1

$\left(i\right)2m=6\Rightarrow m=\frac{6}{2}=3$

Pythagorean triplet is $(m^2-1,2m,m^2+1)$

$=(3^2-1,2\times 3,3^2+1)$

$=(9-1,6,9+1)$

$=(8,6,10)$

$\left(ii\right)2m=14\Rightarrow m=\frac{14}{2}=7$

Pythagorean triplet is $(m^2-1,2m,m^2+1)$

$=(7^2-1,2\times 7,7^2+1)$

$=(49-1,14,49+1)$

$=(48,14,50)$

$\left(iii\right)2m=16\Rightarrow m=\frac{16}{2}=8$

Pythagorean triplet is $(m^2-1,2m,m^2+1)$

$=(8^2-1,2\times 8,8^2+1)$

$=(64-1,16,64+1)$

$=(63,16,65)$