Difference between revisions of "1998 JBMO Problems/Problem 3"

(Created page with "Find all pairs of positive integers <math>(x,y)</math> such that <cmath>x^y = y^{x - y}.</cmath> == Solution == Note that <math>x^y</math> is at least one. Then <math>y^{x -...")
 
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Since <math>c</math> divides the RHS of this equation, it must divide the LHS. Since <math>\gcd(b, c) = 1</math> by assumption, we must have <math>c = 1</math>, so that the equation reduces to <math>b + 1 = a^b - 1</math>, or <math>b + 2 = a^b</math>. This equation has only the solutions <math>b = 1, a = 3</math> and <math>b = 2, a = 2</math>.
 
Since <math>c</math> divides the RHS of this equation, it must divide the LHS. Since <math>\gcd(b, c) = 1</math> by assumption, we must have <math>c = 1</math>, so that the equation reduces to <math>b + 1 = a^b - 1</math>, or <math>b + 2 = a^b</math>. This equation has only the solutions <math>b = 1, a = 3</math> and <math>b = 2, a = 2</math>.
  
Therefore, our only solutions are <math>x = 3^{1 + 1} = 9, y = 3^1 = 3</math>, and <math>x = 2^{2+1} = 8, y = 2^2 = 4</math>, and we are done.
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Therefore, our only solutions are <math>x = 3^{1 + 1} = 9, y = 3^1 = 3</math>, and <math>x = 2^{2+1} = 8, y = 2^1 = 2</math>, and we are done.
  
 
== See also ==
 
== See also ==

Latest revision as of 01:34, 18 September 2020

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.

See also

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