Find a Pythagoras triplet one of whose numbers is 7.

We know that $2m, m^2 + 1, and m^2 - 1$ form a Pythagorean Triplet for any number m, m > 1.

Let 2m = 7

So, m = $\frac{7}{2} = 3.5$

And,  $m^2 + 1$

= $(3.5)^2 + 1$

= $12.25 + 1$

= 13.25

And, $m^2 - 1$

= $(3.5)^2 - 1$

= $12.25 - 1$

= 11.25

Checking if $(7)^2+ (11.25)^2 = (13.25)^2$

⇒$ 49 + 126.5625 = 175.5625$

⇒ $175.5625 = 175.5625$