Question

Write a Pythagorean triplet whose one member is 6.

Answer 1

Hint: In this question, we are given one of the members of the Pythagorean triplet and we have to find the rest two members of the triplet. For this, we will use the basic form of Pythagorean triplet which is $2m,{{m}^{2}}-1\text{ and }{{m}^{2}}+1$ where m can be any natural number. We will compare all three forms by the given number and find the value of m from one of them. After finding the value of m, we will use other forms to find the other two members of the Pythagorean triplet.

Complete step-by-step solution:
Here, we are given one of the members of the triplet as 6.
But we know, $2m,{{m}^{2}}-1\text{ and }{{m}^{2}}+1$ form a Pythagorean triplet. So, let us compare all of them by 6 one by one to get the value of m.
Comparing 6 with 2m, we get:
\[2m=6\Rightarrow m=3\]
Hence, we get the value of m as 3.
Comparing 6 with ${{m}^{2}}-1$ we get:
\[{{m}^{2}}-1=6\Rightarrow {{m}^{2}}=7\]
Since, 7 is not a square number of any integer. Therefore, ${{m}^{2}}-1$ cannot be equal to 6. Thus, no value of m is obtained from ${{m}^{2}}-1$.
Comparing 6 with ${{m}^{2}}+1$ we get:
\[{{m}^{2}}+1=6\Rightarrow {{m}^{2}}=5\]
Since, 5 is not a square number of any integer. Therefore, ${{m}^{2}}+1$ cannot be equal to 5. Thus, no value of m is obtained from ${{m}^{2}}+1$.
Hence, the value of m will only be 3.
Pythagorean triplet formed by taking m=3 are
\[\begin{align}
  & \Rightarrow 2m=23=6 \\
 & \Rightarrow {{m}^{2}}-1={{\left( 3 \right)}^{2}}-1=9-1=8 \\
 & \Rightarrow {{m}^{2}}+1={{\left( 3 \right)}^{2}}+1=9+1=10 \\
\end{align}\]
Hence, (6, 8, 10) form a Pythagorean triplet.

Note: Students should note that, Pythagorean triplet means three numbers which satisfy Pythagoras theorem. They can check their answer by comparing it with Pythagoras theorem. As we can see, ${{\left( 6 \right)}^{2}}+{{\left( 8 \right)}^{2}}={{\left( 10 \right)}^{2}}\Rightarrow 36+64=100$ that implies 100 = 100. Hence, (6, 8, 10) forms a Pythagorean triplet. One more formula which gives us different Pythagorean triplet are $\left( 2n+1,2{{n}^{2}}+2n,2{{n}^{2}}+2n+1 \right)$.