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Old 2008-10-07, 19:36   #1
davar55's Avatar
May 2004
New York City

3·17·83 Posts
Default An Equation to Solve

Here's a curiosity that resembles Fermat's Last:

Solve a^b + b^c = c^a in integers.
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Old 2008-10-07, 22:13   #2
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petrw1's Avatar
Nov 2006
Saskatchewan, Canada

32·5·103 Posts
Default How about

1 1 2

Last fiddled with by petrw1 on 2008-10-07 at 22:16
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Old 2008-10-07, 22:25   #3
cheesehead's Avatar
"Richard B. Woods"
Aug 2002
Wisconsin USA

22×3×641 Posts

0 n 0, for any integer n > 0

n 0 1, for any integer n

0 1 n, for any integer n

Last fiddled with by cheesehead on 2008-10-07 at 22:49
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Old 2008-10-09, 00:35   #4
Kevin's Avatar
Aug 2002
Ann Arbor, MI

433 Posts

Brute force (1<=a,b,c<=1000) makes it appear as though there's no solution in positive integers besides 1,1,2 (and I imagine some inequality trickery could show this to be the case). I'll check on the cases where you include negative numbers later.
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