Wacky diophantine equation

Some preliminaries: the left-hand side is a positive integer; so the right-hand side must also be a positive integer; it follows that $x-y \ge 0$ .

If $x=y$ , we have $x^x = x^0 = 1$ , so the only solution (in positive integers) is $x=y=1$ . From now on we can restrict our search to $x>y$ .

Using this inequality, we have $y^{x-y} = x^y > y^y$ so that $x-y>y$ , or $x>2y$ .

Let $x=ty$ , where $t>2$ from the above inequality. Then the equation becomes $(ty)^y = y^{ty-y} = y^{y(t-1)}$ . Taking the $y^{th}$ root of both sides gives $ty = y^{t-1}$ , or $t=y^{t-2}$ .

We can rearrange this to get $y=t^{\frac{1}{t-2}}$ and $x=t^{\frac{t-1}{t-2}}$ . Any value of $t>2$ will give a solution pair; but which ones give integers?

We'll focus on the expression for $y$ , and treat it as a function of $t$ : $y(t)=t^{\frac{1}{t-2}}$ . First, we'll show that $y(t)$ is decreasing for $t>2$ .

Take logs: $\log{y} = \frac{\log{t}}{t-2}$

Differentiate: $\frac{y'(t)}{y(t)} = \frac{\frac{t-2}{t} - \log{t}}{(t-2)^2}$

Note that $y(t)>0$ , since $t$ is positive; also $(t-2)^2$ is positive. So we are interested in the sign of $\frac{y'(t) (t-2)^2}{y(t)} = \frac{t-2}{t} - \log{t} = 1-\frac{2}{t}-\log{t}$

For $2<t<e$ : $1-\frac{2}{t}-\log{t} < 1-\frac{2}{t}-\log{2} < 1-\frac{2}{e} - \log{2} < 0$

For $t \ge e$ : $1-\frac{2}{t}-\log{t} < 1-\log{t} < 0$

So $y(t)$ is decreasing for $t>2$ . Now, note that $y(3)=3$ and $y(4)=2$ . So there are no solutions for $3<t<4$ (as there are no integers between $3$ and $2$ ). Also, $y(t)>1$ for all $t>2$ ; so there are no solutions for $t>4$ either.

This leaves the region $2<t<3$ . Now, it is possible to find $t$ such that $y(t)$ is an integer; so we need a new strategy.

The last trick is to use the fact that $t=\frac{x}{y}$ is rational, since both $x$ and $y$ are integers. Recall that we found $y(t)=t^{\frac{1}{t-2}}$ . Substituting in for $t$ , this is

$y=\left( \frac{x}{y} \right) ^{\frac{y}{x-2y}}$

$y^{x-2y}=\left( \frac{x}{y} \right) ^y$

The left-hand side of the last expression is an integer. The right-hand side is a rational raised to the power of an integer; the only way for this to also be an integer is if the rational number itself - ie $\frac{x}{y}$ - is an integer too.

So we've shown that for both $x$ and $y$ to be integers, $t=\frac{x}{y}$ must also be an integer. Therefore there are no solutions in the interval $2<t<3$ .

In summary, letting $t=\frac{x}{y}$ :

• for $t=1$ , the only solution is $(x,y)=\boxed{(1,1)}$
• we found that $y(t)=t^{\frac{1}{t-2}}$
• for both $x$ and $y$ to be integers, it is necessary for $t$ to be an integer
• for $t=3$ , we get $(x,y)=\boxed{(9,3)}$
• for $t=4$ , we get $(x,y)=\boxed{(8,2)}$
• for $t>4$ , $1<y(t)<2$ , and so there are no other solutions