Question

Write a pythagorean triplet whose smallest member is
(i) 6
(ii) 14
(iii) 16
(iv) 20

Answer 1

For every number m > 1, the Pythagorean triplet is $\left(2m,m^2-1,m^2+1\right)$.

Using the above result:

(i)
$$\begin{gathered} 2m=6 \\ m=3,m^2=9 \\ {m}^2-1=9-1=8 \\ {m}^2+1=9+1=10 \end{gathered}$$

Thus, the Pythagorean triplet is $\left[6,8,10\right]$.

(ii)
$$\begin{gathered} 2m=14 \\ m=7,m^2=49 \\ {m}^2-1=49-1=48 \\ {m}^2+1=49+1=50 \end{gathered}$$

Thus, the Pythagorean triplet is $\left[14,48,50\right]$.

(iii)
$$\begin{gathered} 2m=16 \\ m=8,m^2=64 \\ {m}^2-1=64-1=63 \\ {m}^2+1=64+1=65 \end{gathered}$$

Thus, the Pythagorean triplet is: $\left[16,63,65\right]$

(iv)
$$\begin{gathered} 2m=20 \\ m=10,m^2=100 \\ {m}^2-1=100-1=99 \\ {m}^2+1=100+1=101 \end{gathered}$$

Thus, the Pythagorean triplet is $\left[20,99,101\right]$.