Question:
If one member of a Pythagorean triplet is $2 \mathrm{~m}$, then the other two members are
(a) $m, m^{2}+1$
(b) $m^{2}+1, m^{2}-1$
(c) $m^{2}, m^{2}-1$
(a) $m, m^{2}+1$
(b) $m^{2}+1, m^{2}-1$
(c) $m^{2}, m^{2}-1$
(d) $m^{2}, m+1$
Solution:
$2 m=4$
$\Rightarrow \quad m=2$
$m^{2}+1=2^{2}+1=4+1=5$
and $\quad m^{2}-1=2^{2}-1=4-1=3$
Now. $\quad 3^{2}+4^{2}=5^{2}$
$\Rightarrow \quad 9+16=25 .$
$\Rightarrow \quad 25=25$
So, 3,4 and 5 are pythagorean triplets.
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