Prove the identity $\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}$ Prove the identity $$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}.$$
We have $$\left(\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}\right)^2=17+2\sqrt{30}-2\sqrt{17+2\sqrt{30}}\cdot\sqrt{17-2\sqrt{30}}+17-2\sqrt{30}=34-2\sqrt{(17)^2-(2\sqrt{30})^2}=34-2\sqrt{289-120}=34-2\sqrt{169}=34-2.13=8=(2\sqrt2)^2, $$
so $$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}.$$
In the hints the authors have written that I should use the fact that the LHS is positive and square it. What would be the problem if it wasn't positive? The identity obviously won't hold because LHS<0, RHS>0...

Since the roots on the LHS get "annihilated", it makes very much sense to look for natural $m,n$ such that
\begin{eqnarray*}
17+2\sqrt{30} & = & \left(\sqrt n + \sqrt m\right)^2 & = & n + m + 2\sqrt{nm}\\
17-2\sqrt{30} & = & \left(\sqrt n - \sqrt m\right)^2 & = & n + m - 2\sqrt{nm}
\end{eqnarray*}
From this approach you get
\begin{eqnarray*}
n+m & = & 17\\
\sqrt{nm} & = & \sqrt{30}
\end{eqnarray*}
Obviously $n = 15, m=2$ fit, hence
$$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}= \sqrt{15}+\sqrt 2 - (\sqrt{15}-\sqrt 2)=2\sqrt{2}$$
