Show that $A=\sqrt{\left|40\sqrt2-57\right|}-\sqrt{\left|40\sqrt2+57\right|}$ is a whole number Show that $A$ is a whole number: $$A=\sqrt{\left|40\sqrt2-57\right|}-\sqrt{\left|40\sqrt2+57\right|}.$$
I don't know if this is necessary, but we can compare $40\sqrt{2}$ and $57$: $$40\sqrt{2}\Diamond57,\\1600\times2\Diamond 3249,\\3200\Diamond3249,\\3200<3249\Rightarrow 40\sqrt{2}<57.$$ Is this actually needed for the solution? So $$A=\sqrt{57-40\sqrt2}-\sqrt{40\sqrt2+57}.$$ What should I do next?

Actually it is not necessary to know if $\;40\sqrt2\;$ is greater or less than $\;57$.
Moreover we do not need to square $\;A\;$ or to solve any system of equations.
The check that the original poster does about the comparison between $\;40\sqrt2\;$ and $\;57\;$ is not necessary, indeed just the same we can remove the absolute value inside the first radical as soon as we get the square of a subtraction.
$\begin{align}
A&=\sqrt{\left|40\sqrt2-57\right|}-\sqrt{\left|40\sqrt2+57\right|}=\\
&=\sqrt{\left|40\sqrt2-32-25\right|}-\sqrt{40\sqrt2+32+25}=\\
&=\sqrt{\left|40\sqrt2-\left(4\sqrt2\right)^2-5^2\right|}-\sqrt{40\sqrt2+\left(4\sqrt2\right)^2+5^2}=\\
&=\sqrt{\left|\left(4\sqrt2\right)^2+5^2-40\sqrt2\right|}-\sqrt{\left(4\sqrt2\right)^2+5^2+40\sqrt2}=\\
&=\sqrt{\left|\left(4\sqrt2-5\right)^2\right|}-\sqrt{\left(4\sqrt2+5\right)^2}=\\
&=\sqrt{\left(4\sqrt2-5\right)^2}-\sqrt{\left(4\sqrt2+5\right)^2}=\\
&=\left|4\sqrt2-5\right|-\left(4\sqrt2+5\right)=\\
&\underset{\color{blue}{\overbrace{\text{but }\;4\sqrt2-5=\sqrt{32}-\sqrt{25}>0}}}{=}\left(4\sqrt2-5\right)-\left(4\sqrt2+5\right)=\\
&=-10\;.
\end{align}$
Hence $A$ is the integer $-10$.
