Question

write a pythagorean triplet whose smallest member is 8

Answer 1

We know Pythagorean triplets are 2m,
${m}^{2} + 1$
and
${m}^{2} - 1$

let 2m = 8
m = 8/2
m = 4

m = 4
so m^2 + 1...
${m}^{2} + 1 = {4}^{2} + 1 = 17$
so m^2 - 1....
${m}^{2} - 1 = {4}^{2} - 1 = 15$
Hence, Pythagorean triplet whose smallest number 8 is 8, 17 and 15
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