Question

If one member of a Pythagorean triplet is $2m$, then the other two members are

• A. $m,m^2+1$
• B. $m^2+1,m^2-1$
• C. $m^2,m^2-1$
• D. $m^2,m+1$

Answer 1

The correct option is B

$m^2+1,m^2-1$

Explanation of the correct option.

Option $\left(B\right)$: $m^2+1,m^2-1$

Since, we know that for natural number $m>1$,

Let $2m,m^2+1,m^2-1$forms a Pythagorean Triplet.

For example let $m=3$

Then the Pythagorean triplet $\left(2m,m^2+1,m^2-1\right)=\left(6,10,8\right)$

$$\begin{gathered} \therefore {10}^2=6^2+8^2 \\ \Rightarrow 100=100 \end{gathered}$$

Hence, Option $\left(\mathrm{B}\right)$ is correct option.

Explanation of the incorrect option.

Option $\left(A\right)$: $m,m^2+1$

Let $2m,m,m^2+1$forms a Pythagorean Triplet.

For example let $m=3$

Then the Pythagorean triplet $\left(2m,m,m^2+1\right)=\left(6,3,10\right)$

$$\begin{gathered} \therefore {10}^2=6^2+3^2 \\ \Rightarrow 100\ne 45 \end{gathered}$$

Hence, Option $\left(A\right)$ is not correct option.

Option $\left(C\right)$: $m^2,m^2-1$

Let $2m,m^2,m^2-1$forms a Pythagorean Triplet.

For example let $m=3$

Then the Pythagorean triplet $\left(2m,m^2,m^2-1\right)=\left(6,9,8\right)$

$$\begin{gathered} \therefore 9^2=6^2+8^2 \\ \Rightarrow 81\ne 100 \end{gathered}$$

Hence, Option $\left(C\right)$ is not correct option.

Option $\left(D\right)$: $m^2,m+1$

Let $2m,m^2,m+1$forms a Pythagorean Triplet.

For example let $m=3$

Then the Pythagorean triplet $\left(2m,m^2,m+1\right)=\left(6,9,4\right)$

$$\begin{gathered} \therefore 9^2=6^2+4^2 \\ \Rightarrow 81\ne 52 \end{gathered}$$

Hence, Option $\left(D\right)$ is not correct option.

Hence, Option $\left(\mathrm{B}\right)$ is the correct option.