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October 4th, 2019, 10:06 AM   #1
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Talking $m^n-n^m=mn$

$m,n \in \mathbb{Z}, \; m^n-n^m=mn$. $m,n=?$
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October 4th, 2019, 10:29 AM   #2
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No solution, unless you adopt a convention that 0^0 = 1 or 0^0 = 0, then 0,0 is the only solution.
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October 4th, 2019, 10:57 AM   #3
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Quote:
Originally Posted by DarnItJimImAnEngineer View Post
No solution, unless you adopt a convention that 0^0 = 1 or 0^0 = 0, then 0,0 is the only solution.
I don't adopt conventions.
Show that there is no solution....
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October 4th, 2019, 11:18 AM   #4
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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 non-integer.
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
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October 4th, 2019, 04:01 PM   #5
SDK
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Quote:
Originally Posted by tahirimanov19 View Post
I don't adopt conventions.
Show that there is no solution....
I'm not sure what this statement means. So let me try to explain what Jim was saying in another way.

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.
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