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

(ii) 14

Answers (1)

D Devendra Khairwa

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.

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