# If (5-sqrt(x))^2 = y-20sqrt(2) where x, y are integers, then what are x and y ?

Jun 3, 2015

Expand and equate the irrational parts to help simplify and find:

$x = 8$ and $y = 33$

#### Explanation:

Expand the left hand side:

${\left(5 - \sqrt{x}\right)}^{2}$

$= {5}^{2} - \left(2 \times 5 \times \sqrt{x}\right) + x$

$= 25 - 10 \sqrt{x} + x$

$= \left(x + 25\right) - 10 \sqrt{x}$

In order to eliminate the irrational $- 20 \sqrt{2}$ term with $x$ and $y$ being integers, we must have:

$- 10 \sqrt{x} = - 20 \sqrt{2}$

Divide both sides by $- 10$ to get:

$\sqrt{x} = 2 \sqrt{2}$

So

$x = {\left(2 \sqrt{2}\right)}^{2} = 8$

$y = x + 25 = 8 + 25 = 33$

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Footnote

Why is it possible to equate the irrational parts like this?

Consider the set of all numbers of the form $a + b \sqrt{2}$ where $a$ and $b$ are rational numbers.

These representations are unique:

Suppose $a + b \sqrt{2} = c + d \sqrt{2}$

Then $a - c = \left(d - b\right) \sqrt{2}$

So if $d \ne b$ we would find $\sqrt{2} = \frac{a - c}{d - b}$ which is a rational number, but $\sqrt{2}$ is irrational.

So we must have $d = b$ and hence $a = c$.