Q.2 Write a Pythagorean triplet whose one member is.

(iii) 16

Answers (1)

For any natural number m > 1, 2m, m^{2} - 1 and m^{2} + 1 forms a Pythagorean triplet.

So if we take,           m^{2} - 1 = 16

                                     m^{2} = 16 + 1 = 17

But the value of m will not be an integer.

   Now we take,          m^{2} + 1 = 16

                                     m^{2} = 16 - 1 = 15

  but the value of m will not be an integer.

If we take                       2m = 16

 then                                 m = 8

Then              m^{2} -1 = 64 - 1 = 63           and          m^{2} +1 =  64 + 1  = 65.

Therefore the required numbers are 16, 63 and 65.

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