1998 JBMO Problems/Problem 3

Find all pairs of positive integers such that

Solution

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

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First Approach (Algebraic Parameterization)

Note that \(x^y \geq 1\), so \(y^{x - y} \geq 1\) as well. This implies \(x \geq y\).

Suppose \(x = a^{b + c}\), \(y = a^c\) for some integer \(a > 0\) and integers \(b, c \geq 1\) with \(\gcd(b, c) = 1\). This ensures that \(x \geq y\).

Then the original equation becomes: \[ (a^{b + c})^{a^c} = (a^c)^{a^{b + c} - a^c} \] Taking logarithms base \(a\), we get: \[ (b + c) \cdot a^c = c \cdot (a^{b + c} - a^c) \] Divide both sides by \(a^c\): \[ b + c = c(a^b - 1) \] Now, since \(\gcd(b, c) = 1\), and \(c\) divides the right-hand side, it must divide the left-hand side. Hence, \(c = 1\). Substituting back: \[ b + 1 = a^b - 1 \Rightarrow a^b = b + 2 \] We now solve for small values of \(b\):

- If \(b = 1\): \(a^1 = 1 + 2 = 3 \Rightarrow a = 3\) - If \(b = 2\): \(a^2 = 2 + 2 = 4 \Rightarrow a = 2\) - Larger \(b\) gives too big values of \(a^b\), so no further solutions.

So the only valid pairs \((a, b)\) are \((3, 1)\) and \((2, 2)\). We now compute:

- For \((a, b) = (3, 1)\), \(c = 1\):

 \(y = a^c = 3\), \(x = a^{b + c} = 3^2 = 9\)

- For \((a, b) = (2, 2)\), \(c = 1\):

 \(y = 2\), \(x = 2^3 = 8\)

So the corresponding solutions are \((x, y) = (9, 3)\) and \((8, 2)\).

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Second Approach (Logarithmic Substitution)

Assume \(x = ky\) for some integer \(k > 1\). Then: \[ (ky)^y = y^{ky - y} = y^{y(k - 1)} \] Taking natural logarithms: \[ y \ln(ky) = y(k - 1)\ln y \Rightarrow \ln(ky) = (k - 1)\ln y \] Expanding the left-hand side: \[ \ln k + \ln y = (k - 1)\ln y \Rightarrow \ln k = (k - 2)\ln y \Rightarrow \ln y = \frac{\ln k}{k - 2} \]

For \(y\) to be an integer, the RHS must be \(\ln\) of an integer.

Try small values of \(k\): - \(k = 3\): \(\ln y = \frac{\ln 3}{1} = \ln 3 \Rightarrow y = 3, x = 9\) - \(k = 4\): \(\ln y = \frac{\ln 4}{2} = \ln 2 \Rightarrow y = 2, x = 8\) - \(k = 5\): \(\ln y = \frac{\ln 5}{3} \notin \ln(\mathbb{Z})\) ⟹ discard

No other values of \(k\) give integer \(y\). Also, checking small \(y\) values directly:

- \(y = 1\): Then \(x^1 = 1^{x - 1} = 1 \Rightarrow x = 1\) ⟹ \((1, 1)\) is a solution.

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Final Answer

The only positive integer solutions to the equation \(x^y = y^{x - y}\) are: \[ \boxed{(x, y) = (1, 1),\ (8, 2),\ (9, 3)} \]

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: \[ \boxed{(x, y) = (1, 1),\ (8, 2),\ (9, 3)} \]

See also

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