An Equation to Solve

davar55 · 2008-10-07, 19:36

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

Solve a^b + b^c = c^a in integers.

petrw1 · 2008-10-07, 22:13

1 1 2

cheesehead · 2008-10-07, 22:25

0 n 0, for any integer n > 0

n 0 1, for any integer n

0 1 n, for any integer n

Kevin · 2008-10-09, 00:35

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.

2008-10-07, 19:36	#1
davar55 May 2004 New York City 4232₁₀ 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 1000111110111₂ 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

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2008-10-09, 00:35	#4
Kevin Aug 2002 Ann Arbor, MI 1B1₁₆ 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.