Question

Write a Pythagorean triplet whose one of the member is 26

Answer 1

$we \: know \\ 2m, \: {m}^{2} + 1, \: {m}^{2} - 1 \\ 2m = 26 \\ m = \frac{26}{2} = 13 \\ \\ {m}^{2} + 1 \\ = {13}^{2} + 1 \\ = 169 + 1 = 170 \\ \\ {m}^{2} - 1 \\ = {13}^{2} + 1 \\ = 169 - 1 = 168 \\ \\ phythagorean \: triplet \: of \: 26 \: are \\ 26, \: 170, \: 168$