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 2008-10-07, 19:36 #1 davar55     May 2004 New York City 5×7×112 Posts An Equation to Solve Here's a curiosity that resembles Fermat's Last: Solve a^b + b^c = c^a in integers.
 2008-10-07, 22:13 #2 petrw1 1976 Toyota Corona years forever!     "Wayne" Nov 2006 Saskatchewan, Canada 3×37×47 Posts How about 1 1 2 Last fiddled with by petrw1 on 2008-10-07 at 22:16
 2008-10-07, 22:25 #3 cheesehead     "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
 2008-10-09, 00:35 #4 Kevin     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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