Question

Write a Pythagorean triplet whose one member is 14.

Answer 1

There are three numbers $2m,m^2-1and{m}^2+1,$ in a Pythagorean Triplet $(for,m>1)$.

$Here,2m=14\Rightarrow m=\frac{14}{2}=7\therefore \text{Second number}(m^2-1)=(7{)}^2-1=49-1=48\text{And third number}(m^2+1)=(7{)}^2+1=49+1=50$
Hence, the Pythagorean triplet is (14, 48, 50).