Question

Find the other two numbers of the Pythagorean triplet in which 28 is the smallest number.

• A. 14, 15
• B. 195, 197
• C. 196, 197
• D. 195, 196

Answer 1

The correct option is B 195, 197

We know that for even numbers, Pythagorean triplets can be represented as

$2m,(m^2-1)and(m^2+1)$

$\Rightarrow (2m{)}^2+(m^2-1{)}^2=(m^2+1{)}^2$

Here, $2m=28$.

$\Rightarrow m=14$

$\Rightarrow m^2={14}^2=196$

$\Rightarrow m^2-1=195$

$\Rightarrow m^2+1=197$

Thus, the required Pythagorean triplet is 28, 195 and 197.