20081007, 19:36  #1 
May 2004
New York City
5·7·11^{2} Posts 
An Equation to Solve
Here's a curiosity that resembles Fermat's Last:
Solve a^b + b^c = c^a in integers. 
20081007, 22:13  #2 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
1381_{16} Posts 
How about
1 1 2
Last fiddled with by petrw1 on 20081007 at 22:16 
20081007, 22:25  #3 
"Richard B. Woods"
Aug 2002
Wisconsin USA
1E0C_{16} 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 20081007 at 22:49 
20081009, 00:35  #4 
Aug 2002
Ann Arbor, MI
110110001_{2} 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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