# Side lengths of an acute triangle are sqrtn, sqrt(n+1), and sqrt(n+2). How do you find n?

${\left(\sqrt{n + 2}\right)}^{2} < {\left(\sqrt{n}\right)}^{2} + {\left(\sqrt{n + 1}\right)}^{2}$
$\implies n + 2 < n + n + 1$
$\implies n > 1$