A triangle has sides $2n, n^2+1$ and $n^2-1$ prove that it is right angled

Question

I've tried using Pythagoras theorem but it always results in a silly answer like $n=n^2$ or something.

I'm nearly 100% sure this is done with Pythagoras but I'm not sure which way to do it

Answer 1

Hint: $$ 4n^2 = (n^4 + 2n^2 +1) - (n^4 - 2n^2 + 1)$$

Answer 2

This is a particular case of Pythagorean triples. All such triples $(a,b,c)$ with $a^2+b^2=c^2$ can be generated by two integers $m,n$ as: $$ a=n^2-m^2,\qquad b=2mn,\qquad c=m^2+n^2 $$ (the identy $a^2+b^2=c^2$ is easely verified). In your case we have $m=1$.