If the Pythagorean triples of one member is $22$ , find the other two members.

124

and

122

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123

and

122

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121

and

122

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120

and

122

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Solution

The correct option is D $120$ and $122$
As we know $2 m$ , $m^{2} + 1, m^{2} - 1$ form a Pythagorean triplet for any number $m > 1$ .

Let us assume $2 m = 22$

$\Rightarrow m = 11$

Therefore, $m^{2} + 1$ $=$ $11^{2} + 1$

$\Rightarrow 121 + 1 = 122$

and $m^{2} - 1$ $=$ $11^{2} - 1$

$\Rightarrow 121 - 1 = 120$

Hence, the other two members are $120$ and $122$ .

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