Question

Which of the following triplets are Pythagorean?
(i) $ \left( {16,63,65} \right) $ (ii) $ \left( {24,144,145} \right) $

Answer 1

Hint: We know that for three numbers to be a Pythagorean triplet, they should be in the form $ 2m $ , $ {m^2} - 1 $ and $ {m^2} + 1 $ . We will check for the given triplets whether they satisfy the condition of Pythagorean triplets or not.

Complete step-by-step answer:
(i)
The given triplets are $ \left( {16,63,65} \right) $ .
Let us assume 16 as the smallest even number, then it should be equivalent to $ 2m $ .
Then we can write
 $ \begin{array}{c}
2m = 16\\
m = 8
\end{array} $
 Then from this we can find the other two numbers from the expression $ {m^2} - 1 $ and $ {m^2} + 1 $ by substituting 8 for $ m $ .This can be expressed as:
\[\begin{array}{c}
{m^2} - 1 = {8^2} - 1\\
 = 64 - 1\\
 = 63
\end{array}\]
And
\[\begin{array}{c}
{m^2} + 1 = {8^2} + 1\\
 = 64 + 1\\
 = 65
\end{array}\]
From the above result we can say that triples $ \left( {16,63,65} \right) $ is a Pythagorean triplet.
(i)
The given triplets are $ \left( {24,144,145} \right) $ .
Let us assume 24 as the smallest even number, then it should be equivalent to $ 2m $ .
Then we can write
 $ \begin{array}{c}
2m = 24\\
m = 12
\end{array} $
 Then from this, we can find the other two numbers from the expression $ {m^2} - 1 $ and $ {m^2} + 1 $ by substituting 12 for $ m $ .This can be expressed as:
\[\begin{array}{c}
{m^2} - 1 = {12^2} - 1\\
 = 144 - 1\\
 = 143
\end{array}\]
And
\[\begin{array}{c}
{m^2} + 1 = {12^2} + 1\\
 = 144 + 1\\
 = 145
\end{array}\]
From the above result we can say that triples $ \left( {24,144,145} \right) $ is not a Pythagorean triplet since it does not satisfy the condition of Pythagorean triplets.

Note: If the three numbers say $ \left( {a,b,c} \right) $ are the Pythagorean triplets, then we can say that $ \left( {ka,kb,kc} \right) $ will also be Pythagorean triplets. If the number considered is odd, then for finding the Pythagorean triplet, we will have to consider triplets as $ 2{m^2} $ , \[\dfrac{{{m^2} - 1}}{2}\] and \[\dfrac{{{m^2} + 1}}{2}\] . The condition depends upon the question. In this question, we are considering the smallest even number as $ 2m $ .