October 21st, 2012, 05:24 AM  #1 
Newbie Joined: Oct 2012 Posts: 3 Thanks: 0  Solutions of a^xa=b^yb
While trying to solve a problem concerning natural numbers, i struck upon this equation: a^xa=b^yb (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^xa=b^yb
I can't answer the question completely, but: Assume, both greater than 1. can thus be rewritten and rearranged as . We can, without loss of generality, as take (as is a given) , with with , so we can rewrite this as , which can be rewritten as . As b,d,x are all natural numbers, there is a binomial expansion of , the first term of which is . Taking all of the other terms of the expansion as Q, we get , or simply . As Q is the binominal expansion of 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 , 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^xa=b^yb
That looks like a good start. Thus, we can conclude that with (the set of the prime numbers) and and not dividable by nor is impossible for . But what if ? 
October 22nd, 2012, 06:11 AM  #4 
Newbie Joined: Oct 2012 Posts: 3 Thanks: 0  Re: Solutions of a^xa=b^yb
exept if 
October 25th, 2012, 03:47 AM  #5 
Member Joined: Oct 2009 From: Northern Virginia Posts: 31 Thanks: 1  Re: Solutions of a^xa=b^yb
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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