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 October 21st, 2012, 05:24 AM #1 Newbie   Joined: Oct 2012 Posts: 3 Thanks: 0 Solutions of a^x-a=b^y-b While trying to solve a problem concerning natural numbers, i struck upon this equation: a^x-a=b^y-b (with a, b, x and y natural numbers and a not equal to b) My question is: For wich values of a and b has this equation solutions?
 October 21st, 2012, 05:21 PM #2 Member   Joined: Oct 2009 From: Northern Virginia Posts: 31 Thanks: 1 Re: Solutions of a^x-a=b^y-b I can't answer the question completely, but: Assume$x= y$, both greater than 1. $a^x-a=b^y-b$ can thus be rewritten and rearranged as $a^x - b^x= a - b$. We can, without loss of generality, as take $a > b$ (as $a \neq b$ is a given) , with $d= a - b$ with $d > 1$ , so we can rewrite this as $(b + d)^x - b^x= b + d - b$, which can be rewritten as $(b + d)^x= d$. As b,d,x are all natural numbers, there is a binomial expansion of $(b + d)^x$, the first term of which is $b^x$. Taking all of the other terms of the expansion as Q, we get $b^x + Q - b^x= d$, or simply $Q= d$. As Q is the binominal expansion of $(b + d)^x$ minus the first term, every term in Q has a factor of d. Thus Q can be written dQ', where Q' = Q/d, and Q' is an integer. So the equation becomes Q' = 1. But this appears impossible for any natural numbers a, b, x, with x > 1, because at least the first remaining term of the expansion has a factor $\binom{x}{1}= x$, which is larger than one, and all the terms being added are positive. Thus there are no solutions with x = y both greater than 1.
 October 22nd, 2012, 05:37 AM #3 Newbie   Joined: Oct 2012 Posts: 3 Thanks: 0 Re: Solutions of a^x-a=b^y-b That looks like a good start. Thus, we can conclude that $\left\{ \begin{array}{llll} n-p_1 = p_2^{\alpha}\\ n-p_2 = p_1^{\beta}\\ \end{array} \right$ with $p_1,\;p_2 \in \mathbb{P}$ (the set of the prime numbers) and $n,\;\alpha,\;\beta \in \mathbb{N}$ and $n$ not dividable by $p_1$ nor $p_2$ is impossible for $\alpha=\beta$. But what if $\alpha \neq \beta$?
 October 22nd, 2012, 06:11 AM #4 Newbie   Joined: Oct 2012 Posts: 3 Thanks: 0 Re: Solutions of a^x-a=b^y-b exept if $\alpha= \beta = 1$
 October 25th, 2012, 03:47 AM #5 Member   Joined: Oct 2009 From: Northern Virginia Posts: 31 Thanks: 1 Re: Solutions of a^x-a=b^y-b a^x - a = b^y - b or n = a^x - a n = b^y - b It is obvious that n is divisible by both a and b. It is also clear that n is divisible by 2. I wrote a program to search for solutions using a brute force method, and found these: 3^2 - 3 = 2^3 - 2 = 6 6^2 - 6 = 2^5 - 2 = 30 13^3 - 13 = 3^7 - 3 = 2184 15^2 - 15 = 6^3 - 6 = 210 16^2 - 16 = 3^5 - 3 = 240 91^2 - 91 = 2^13 - 2 = 8190 280^2 - 280 = 5^7 - 5 = 78120 4930^2 - 4930 = 30^5 - 30 = 24299970 I note that, of the solutions I found, all of the exponents are prime (2, 3, 5, 7, 13), although the a"s and b"s are not. In the cases such that I found where both the exponents are prime, by Fermat's Little Theorem n is also divisible by x and y. I don't see any other obvious patterns. Both a and b can be odd (a = 13, b = 3), both even (a = 6, b = 2) or one even and one odd ( a = 15, b = 6; a = 15, b = 3). a and b can both be prime, or both composite, or one of each.

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