1998 JBMO Problems/Problem 3

Find all pairs of positive integers $(x,y)$ such that $x^y = y^{x - y}.$

Solution

Note that $x^y$ is at least one. Then $y^{x - y}$ is at least one, so $x \geq y$ .

Write $x = a^{b+c}, y = a^c$ , where $\gcd(b, c) = 1$ . (We know that $b$ is nonnegative because $x\geq y$ .) Then our equation becomes $a^{(b+c)*a^c} = a^{c*(a^{b+c} - a^c)}$ . Taking logarithms base $a$ and dividing through by $a^c$ , we obtain $b + c = c*(a^b - 1)$ .

Since $c$ divides the RHS of this equation, it must divide the LHS. Since $\gcd(b, c) = 1$ by assumption, we must have $c = 1$ , so that the equation reduces to $b + 1 = a^b - 1$ , or $b + 2 = a^b$ . This equation has only the solutions $b = 1, a = 3$ and $b = 2, a = 2$ .

Therefore, our only solutions are $x = 3^{1 + 1} = 9, y = 3^1 = 3$ , and $x = 2^{2+1} = 8, y = 2^1 = 2$ , and we are done.

We are given the equation: \[ x^y = y^{x - y} \] and asked to find all positive integers $(x, y)$ that satisfy it.

\textbf{Step 1: Try small values of $y$}

We begin by checking small values of $y$:

- If $y = 1$, then the equation becomes:

 \[
 x^1 = 1^{x - 1} = 1 \Rightarrow x = 1
 \]
 So \((x, y) = (1, 1)\) is a solution.

- If $y = 2$, try $x = 8$:

 \[
 x^y = 8^2 = 64,\quad y^{x - y} = 2^6 = 64
 \]
 So \((x, y) = (8, 2)\) is a solution.

- If $y = 3$, try $x = 9$:

 \[
 x^y = 9^3 = 729,\quad y^{x - y} = 3^6 = 729
 \]
 So \((x, y) = (9, 3)\) is a solution.

\textbf{Step 2: General approach using logarithms}

Assume $x = ky$ for some integer $k > 1$. Then the equation becomes: \[ (ky)^y = y^{ky - y} = y^{y(k - 1)} \] Taking natural logarithms: \[ y \ln(ky) = y(k - 1)\ln y \Rightarrow \ln(ky) = (k - 1)\ln y \] Expanding the left-hand side: \[ \ln k + \ln y = (k - 1)\ln y \Rightarrow \ln k = (k - 2)\ln y \Rightarrow \ln y = \frac{\ln k}{k - 2} \] So for $y$ to be an integer, $\frac{\ln k}{k - 2}$ must be the logarithm of an integer.

Try small values of $k$: - If $k = 3$: $\ln y = \frac{\ln 3}{1} = \ln 3 \Rightarrow y = 3, x = 9$ - If $k = 4$: $\ln y = \frac{\ln 4}{2} = \ln 2 \Rightarrow y = 2, x = 8$ - If $k = 5$: $\ln y = \frac{\ln 5}{3} \not\in \ln(\mathbb{Z})$, so no integer solution for $y$

No other values of $k$ give integer solutions for $y$.

\textbf{Final Answer:} The only positive integer solutions are: \[ \boxed{(x, y) = (1, 1),\ (8, 2),\ (9, 3)} \]

1998 JBMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
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1998 JBMO Problems/Problem 3

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