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 \[\begin{equation} x-2\sqrt{11x}+11=y. \label{eq:1} \end{equation}\]

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).