Question

Let m ∈ N. (a) Prove that m2 – 1, 2m, m2 + 1 is a Pythagorean triple.

Answer 1

${\huge {\mathfrak {Answer :-}}}$

➡️ (m^2 + 1)^2 = (2m)^2 + (m^2 - 1)^2

Or, (m^2 + 1)^2 - (m^2 - 1)^2 = 4m^2

Or, 4 * m^2 * 1 = 4m^2

Or, 4m^2 = 4m^2.

➡️ Hence, verified.

That's it..

Answer 2

$\boxed{\huge{\bold{\blue{Answer}}}}$

♞(m^2 + 1)^2 = (2m)^2 + (m^2 - 1)^2

Or, (m^2 + 1)^2 - (m^2 - 1)^2 = 4m^2

Or, 4 * m^2 * 1 = 4m^2

Or, 4m^2 = 4m^2.

♞Hence, verified.

$\boxed{\huge{\bold{\blue{SoUMoK}}}}$

The above one is 100% correct❕✌✌