Question:
For every natural number $m,\left(2 m-1,2 m^{3}-2 m, 2 m^{2}-2 m+1\right)$ is a pythagorean triplet.
Solution:
False
$\because$ $(2 m-1)^{2} \neq\left(2 m^{3}-2 m\right)^{2}+\left(2 m^{2}-2 m+1\right)^{2}$
$\left(2 m^{3}-2 m\right)^{2} \neq(2 m-1)^{2}+\left(2 m^{2}-2 m+1\right)^{2}$
and $\left(2 m^{2}-2 m+1\right)^{2} \neq(2 m-1)^{2}+\left(2 m^{3}-2 m\right)^{2}$
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