Among the following options which value of a and b satisfies the equation:
$15^{2}$ + $a^{2}$ = $b^{2}$

8, 9

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8, 16

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8, 17

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9, 16

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Solution

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

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