Question
Find a Pythagorean triplet in which one member is $12.$
✓
Answer
If we take $m^2– 1 = 12$
Then,$ m^2= 12 + 1 = 13$
Then the value of $m$ will not be an integer.
So, we try to take $m^2+ 1 = 12$. Again $m^2= 11$ will not give an integer value for $m.$
So, let us take $2m = 12$
then $m = 6$
Thus, $m^2– 1 = 36 – 1 = 35$ and $m^2+ 1 = 36 + 1 = 37$
Therefore, the required triplet is $12, 35, 37.$
Then,$ m^2= 12 + 1 = 13$
Then the value of $m$ will not be an integer.
So, we try to take $m^2+ 1 = 12$. Again $m^2= 11$ will not give an integer value for $m.$
So, let us take $2m = 12$
then $m = 6$
Thus, $m^2– 1 = 36 – 1 = 35$ and $m^2+ 1 = 36 + 1 = 37$
Therefore, the required triplet is $12, 35, 37.$
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