Find a Pythagorean triplet whose one member is 18.

We know that the Pythagorean triplets can be found by using the general form 2m, m² − 1, m² + 1, where m is a natural number greater than 1.

If m² − 1 = 18

⇒ m² = 18 + 1 = 19

Then, the value of m will not be an integer.

If m² + 1 = 18

⇒ m² = 18 − 1 = 17

Then, the value of m will not be an integer.

Therefore, let 2m = 18

⇒ m = 9, which is a natural number greater than 1

∴ m² − 1 = 9² − 1 = 81 − 1 = 80,

m² + 1 = 9² + 1 = 81 + 1 = 82.

Therefore, the required Pythagorean triplet is 18, 80 and 82.