HP Prime & HP 49G Problem with square roots

[Manolo Sobrino]

OK, first let's notice that 1201 is prime, 42035=35*1201 and 1236=35+1201. Now rewrite the longer expression:

\begin{equation} \sqrt{35+2\sqrt{6}}+\sqrt{1201+35-2\sqrt{6}+2\sqrt{\left(35-2\sqrt{6}\right)1201}}
\end{equation}
That is the square of a sum:\begin{equation} \sqrt{35+2\sqrt{6}}+\sqrt{\left(\sqrt{1201}+\sqrt{35-2\sqrt{6}}\right)^2}
\end{equation}
You don't need to worry about the absolute value, it's simply:\begin{equation}\sqrt{1201}+\sqrt{35+2\sqrt{6}}+\sqrt{35-2\sqrt{6}}
\end{equation}
If a>b it's trivial to prove that:\begin{equation}\sqrt{a+b}+\sqrt{a-b}=\sqrt{2a+2\sqrt{a^2-b^2}}\end{equation}
In this case: \begin{equation}\sqrt{70+2\sqrt{1225-4\cdot 6}}=\sqrt{70+2\sqrt{1201}}\end{equation}
There you go.

(You guys should use paper and pencil more often )