Question

Find a and b from below ${15}^2$ + $a^2$ = $b^2$

• A. 8,9
• B. 8,16
• C. 8,17
• D. 9,16

Answer 1

The correct option is C

8,17

$m^2$ - 1, 2m, $m^2$ + 1 form a Pythagorean triplet

As the given number (15) is odd, it can't be of the form 2m.

$m^2$ - 1 = 15

$m^2$ = 16 → 16 is a perfect square.
So, m = 4

2m = 2 x 4 = 8

$m^2$ + 1 = 4² + 1 = 16 + 1 = 17

So, a = 8
and b = 17