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\begin{document}
\title{Is $4\sqrt{4-2\sqrt{3}}+\sqrt{97-56\sqrt{3}}$ an integer?}

\maketitle
That's a question I've found in YouTube's homepage, where I can find
a lot of math videos. I do not want to watch that video because of
ads that interrupt the short lecture's flow. Besides, I know how to
deal with a square root inside a square root. If you don't see that
$4-2\sqrt{3}=1-2\sqrt{3}+3=(1-\sqrt{3})^{2}$, you can find $a,b$
such that:

1. $(a+b\sqrt{3})^{2}=4-2\sqrt{3}$

2. a,b are not multiples of $\sqrt{3}$\\

Let us expand:

$a^{2}+2ab\sqrt{3}+3b^{2}=4-2\sqrt{3}\Rightarrow$

$\Rightarrow\begin{cases}
a^{2}+b^{2} & =4\\
2ab & =-2
\end{cases}$\\

Now, from the 2nd equation:

$b=-\frac{1}{a}$\\

Let us plug it into the 1st equation:

$a^{2}+\frac{3}{a^{2}}=4\Rightarrow$

$\Rightarrow a^{4}+3=4a^{2}\Rightarrow$

$\Rightarrow a^{4}-4a^{2}+3=0\Rightarrow$

$\Rightarrow a^{4}-4a^{2}+4-4+3=0\Rightarrow$

$\Rightarrow(a^{2}-2)^{2}-1=0\Rightarrow$

$\Rightarrow(a^{2}-1)(a^{2}-3)=0$\\

The solution $a^{2}=3$ is unwanted because then $a=\pm\sqrt{3},$so
let us take $a^{2}=1\Rightarrow a=\pm1$.

Let us now determine if $a$ is positive or negative:

$a=1\Rightarrow b=-\frac{1}{a}=-1\Rightarrow a+\sqrt{3}b=1-\sqrt{3}<0$;
this cannot be a square root, so we'll take:

$a=-1\Rightarrow b=1\Rightarrow a+\sqrt{3}b=-1+\sqrt{3}>0$ Great!\\

Now, 

$4\sqrt{4-2\sqrt{3}}=-4+4\sqrt{3}$\\

So, now let us check if there exists $c\in\mathbb{Z}$, such that:

1. $(c-4\sqrt{3})^{2}=97-56\sqrt{3}$

2. $c-4\sqrt{3}>0$\\

OK, let us solve (1):

$c^{2}-8c\sqrt{3}+48=97-56\sqrt{3}\Rightarrow$

$\Rightarrow c^{2}-8c\sqrt{3}=49-56\sqrt{3}\Rightarrow$\\

If we take $c=7$, we'll get:

$\begin{cases}
c^{2} & =49\\
-8c\sqrt{3} & =-56\sqrt{3}
\end{cases}\Rightarrow$

$\Rightarrow c^{2}-8c\sqrt{3}=49-56\sqrt{3}$\\

This solves $(c-4\sqrt{3})^{2}=97-56\sqrt{3}$

and $7-4\sqrt{3}>0$ because $7^{2}=49>48=(4\sqrt{3})^{2}$\\

Now, if we add the 2 square roots:

$4\sqrt{4-2\sqrt{3}}+\sqrt{97-56\sqrt{3}}=-4+4\sqrt{3}+7-4\sqrt{3}=3$

\textbf{\textcolor{red}{An integer!}}
\end{document}