Show that $(m^{2} - 1), (2 m), m^{2} + 1$ always form a pythagoran triplet.

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Solution

Let $a = m^{2} - 1 b = 2 m c = m^{2} + 1$
using pythagoras theorem, $a^{2} + b^{2} = c^{2}$
So, ${(m^{2} - 1)}^{2} + 4 m^{2} = {(m^{2} + 1)}^{2}$
$m^{4} + 1 - 2 m^{2} + 4 m^{2} = {(m^{2} + 1)}^{2} m^{4} + 2 m^{2} + 1 = {(m^{2} + 1)}^{2} {(m^{2} + 1)}^{2} = {(m^{2} + 1)}^{2}$
Since $L H S = R H S$ , we can say that $a, b, c$ always form pythagorean triplet.

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