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...
 A: both of your original real numbers are roots of
$$  x^4 - 34 x^2 + 169  $$
Standard bit for quartic with no cubic term and no linear,
$$  (x^2 - 13)^2 - 8 x^2 =  x^4 - 34 x^2 + 169 $$
$$  (x^2 - 13)^2 -  (x \sqrt 8)^2 =  x^4 - 34 x^2 + 169 $$
This becomes ( because a difference of squares)
$$ (x^2 - x \sqrt 8 - 13)(x^2 + x \sqrt 8 - 13) $$
so that your numbersare two out of four values
$$ \frac{\pm \sqrt 8 \pm \sqrt {60}}{2}   $$
or
$$ \pm \sqrt 2 \pm \sqrt {15}  $$
This leads to checking
$$ (\sqrt 2 + \sqrt {15} )^2 = 17 + 2 \sqrt{30}  $$
along with
$$ (-\sqrt 2 + \sqrt {15} )^2 = 17 - 2 \sqrt{30}  $$
Your original expression is equal to
$$   (\sqrt 2 + \sqrt {15} ) -  (-\sqrt 2 + \sqrt {15} )   $$
A: 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
$$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}= \sqrt{15}+\sqrt 2 - (\sqrt{15}-\sqrt 2)=2\sqrt{2}$$
