October 4th, 2019, 10:06 AM  #1 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle  $m^nn^m=mn$
$m,n \in \mathbb{Z}, \; m^nn^m=mn$. $m,n=?$

October 4th, 2019, 10:29 AM  #2 
Senior Member Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 
No solution, unless you adopt a convention that 0^0 = 1 or 0^0 = 0, then 0,0 is the only solution.

October 4th, 2019, 10:57 AM  #3 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle  
October 4th, 2019, 11:18 AM  #4 
Senior Member Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 
For equal values, LHS is zero (or undefined). For large, unequal positive values, the LHS is going to blow up faster than the RHS. For large, unequal negative values or large negative and positive values, the LHS is going to shrink and become a noninteger. Any solutions would be small numbers. I tried 400 million small number pairs, and the only one that came close was (0,0), because MATLAB thinks 0^0 is 1 cute little birdie. Runners up: (1)^(2)  (2)^(1)  (1)(2) = 1/2 (2)^(2)  (2)^(2)  (2)(1) = 1/4 
October 4th, 2019, 04:01 PM  #5  
Senior Member Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
Suppose I claim that $0^0 = 1$. Then $m = 0 = n$ is a solution to your equation. If you are now claiming there aren't any solutions, then you are adopting the convention that $0^0$ is undefined (or at least not equal to 1 or 0) whether you like it or not.  