Review question

# If $\sqrt{x}-\sqrt{11}=\sqrt{y}$, when is $x/y$ a maximum? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9499

## Solution

Positive integers $x$ and $y$ satisfy the equation $\sqrt{x}-\sqrt{11}=\sqrt{y}$.

What is the maximum possible value of $\dfrac{x}{y}$?

Squaring the equation $\sqrt{x}-\sqrt{11}=\sqrt{y}$ gives $$$x-2\sqrt{11x}+11=y. \label{eq:1}$$$

As $x$ and $y$ are integers, this implies that $2\sqrt{11x}$ is also an integer, and so $11x$ must be a perfect square.

This can only be the case if $x=11a^2$, where $a>0$.

Substituting the expression for $x$ into $\eqref{eq:1}$ gives $y=11a^2-22a+11=11(a^2-2a+1)=11(a-1)^2.$

We now have $a>1,$ as $y>0$, and now the quotient $\dfrac{x}{y}=\left(\dfrac{a}{a-1}\right)^2$ is well-defined (since we are not dividing by $0$).

We have $\dfrac{a}{a-1} = \dfrac{(a-1) + 1}{a-1} = 1 + \dfrac{1}{a-1}$, so as $a$ gets bigger, $\dfrac{a}{a-1}$ gets smaller.

So the maximum value of $\dfrac{a}{a-1}$ happens when $a$ takes its smallest possible value, which is $2$.

This gives $x = 44, y = 11, \dfrac{x}{y} = 4$.

It’s wise to check this solution satisfies the original equations (which it does).