We are looking for all integer solutions for the equation $a^b+1=b^a$. We conjecture that there are only the solutions $(0,b),(1,2),(2,3)$. It is easy to see, that if a is odd and b even, there is only the solution $(1,2)$, but we don't see the general case. We think there should be a simple argument.

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    $\begingroup$ This is a special case of Catalan's equation, solved by Mihailescu. $\endgroup$ Commented Nov 30, 2012 at 18:16
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    $\begingroup$ If $a>b\ge3$, then $b^a-a^b=b^b(b^{a-b}-(1+(a-b)/b)^b)>b^b(b^{a-b}-e^{a-b})>9(3-e)>1$. Similarly, $b>a\ge3$ implies $b^a-a^b< 0$. Thus at least one of $a,b$ must be 0, 1, or 2, and these cases are easy to do by hand. $\endgroup$ Commented Nov 30, 2012 at 18:43
  • $\begingroup$ Assume 0<a<b, write b as a+d, and note the ratio is close to 1 and is also (f/a)^d, where f is a number between 1 and 3. This gives an upper bound on a. Gerhard "Ask Me About System Design" Paseman, 2012.11.30 $\endgroup$ Commented Nov 30, 2012 at 18:47
  • $\begingroup$ Yes, much easier than quoting Mihailescu's theorem! $\endgroup$ Commented Nov 30, 2012 at 23:55


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