Show that $A=\sqrt{\left|40\sqrt2-57\right|}-\sqrt{\left|40\sqrt2+57\right|}$ is a whole number

Question

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?

Answer 1

$$A=\sqrt{57-40\sqrt2}-\sqrt{40\sqrt2+57} = \sqrt{(4\sqrt2-5)^2} - \sqrt{(4\sqrt2+5)^2} $$

$$ = (4\sqrt2-5) - (4\sqrt2+5) = -10$$

So $A$ is the integer $-10$.
It's just written in some slightly convoluted form.

Answer 2

That number is $-10$. In fact, if you try to express $\sqrt{57-40\sqrt2}$ as $a+b\sqrt2$, you solve the system$$\left\{\begin{array}{l}a^2+2b^2=57\\2ab=-40.\end{array}\right.$$You will get that $a=5$ and that $b=-4$. By the same method, you will get that $\sqrt{57+40\sqrt2}=-5+4\sqrt2$. So$$\sqrt{57-40\sqrt2}-\sqrt{57+40\sqrt2}=-5+4\sqrt2-\left(5+4\sqrt2\right)=-10.$$

Answer 3

First of all is $40\sqrt 2$ greater or less than $57$? One assumes it's a little less.

$(\frac {57}{40})^2 = \frac {(60-3)^2}{1600} = \frac {3600 +9 -2\cdot 3\cdot 60}{1600}= \frac {3600-360+ 9}{1600}= \frac {3249}{1600} > 2$ so $40\sqrt 2 < 57$.

Well if you square $A$ you get

$|40\sqrt 2 + 57| + |40\sqrt 2 - 57|-2\sqrt(|40\sqrt 2+57||40\sqrt 2-57|)=$

$(57+ 40\sqrt 2) + (57 - 40 \sqrt 2) - 2\sqrt{(57+40\sqrt 2)(57-40\sqrt 2)}=$

$57 + 57 - 2\sqrt{57^2-(40\sqrt 2)^2}=$

$114 - 2\sqrt{57^2 - 1600\cdot 2}=$

$114 -2 \sqrt {3249 - 3200} =$

$114 - 2\sqrt{49} = $

$114 - 2\cdot 7= $

$114 - 14= 100$.

So $A^2 = 100$ and $A = \pm 10$.

How cute.

.....

Oh, how clever....$57 \pm 40\sqrt 2 = 57 \pm 2\times 20\sqrt 2= 57 \pm 2\times a(b\sqrt 2)$ where $ab = 20$.

To solve $a^2 + 2b^2 = 57$ we have $57 = 25 + 2*16$ and so $57\pm 2\times 40 \sqrt 2 = 5^2 \pm 2\times 5\cdot 4\sqrt 2 + (4\sqrt 2)^2 = (5 \pm 4\sqrt 2)^2$.

So the who thing works out.

$A =(4\sqrt 2 -5) -(4\sqrt 2 + 5) = -10$.

Cute.

Answer 4

The check you're doing is indeed necessary in order to remove the absolute value.

However, when you arrive at $3200\mathrel{\Diamond}3249$ you can realize that the difference is a square, which is precisely the condition for a radical of the form $$ \sqrt{a\pm\sqrt{b}} $$ can be “denested”. Let's see why. Suppose $\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$; after squaring we get $$ a+\sqrt{b}=x+y+\sqrt{4xy} $$ and if we equate the two parts we get $$ \begin{cases} x+y=a \\[6px] 4xy=b \end{cases} $$ Hence $(x-y)^2=(x+y)^2-4xy=a^2-b$. So, in order to find the integers $x,y$, we need that $a^2-b$ is a square. Once we have it, we can determine $x$ and $y$.

Note that the same holds for $\sqrt{a-\sqrt{b}}=\sqrt{x}-\sqrt{y}$.

In your case $a=57,b=3200$ and $a^2-b=49=7^2$, so you get $$ x+y=57,\quad x-y=7 $$ and therefore $x=32,y=25$, from which $$ \sqrt{57\pm40\sqrt{2}}=\sqrt{32}\pm5 $$ and you can finish.

Answer 5

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 $-10$.