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

(ii) 14

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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 = 14

                                     m^{2} = 14 + 1 = 15

But then the value of m will not be an integer.

We take,          m^{2} + 1 = 14

                                     m^{2} = 14 - 1 = 13

but the value of m will not be an integer.

If we take                       2m = 14

or                                     m = 7

Then              m^{2} -1 = 49 - 1 = 48           and          m^{2} +1 =  49 + 1  = 50.

Therefore the combination of number is 14, 48 and 50.

Posted by

Devendra Khairwa

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