Difference between revisions of "1998 JBMO Problems/Problem 3"
(→Solution) |
(→Solution) |
||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
− | + | We are given the equation: | |
+ | \[ | ||
+ | x^y = y^{x - y} | ||
+ | \] | ||
+ | and asked to find all positive integers \((x, y)\) satisfying it. | ||
− | + | --- | |
− | + | == 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 \((8, 2)\). | ||
+ | |||
+ | --- | ||
+ | |||
+ | == 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. | ||
+ | |||
+ | --- | ||
+ | |||
+ | == 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: | We are given the equation: |
Latest revision as of 13:39, 4 July 2025
Find all pairs of positive integers such that
Contents
Solution
We are given the equation: \[ x^y = y^{x - y} \] and asked to find all positive integers \((x, y)\) satisfying it.
---
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 \((8, 2)\).
---
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.
---
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 |