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October 21st, 2012, 05:24 AM   #1
pKs
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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?
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October 21st, 2012, 05:21 PM   #2
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Re: Solutions of a^x-a=b^y-b

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.
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October 22nd, 2012, 05:37 AM   #3
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Re: Solutions of a^x-a=b^y-b

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 ?
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October 22nd, 2012, 06:11 AM   #4
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Re: Solutions of a^x-a=b^y-b

exept if
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October 25th, 2012, 03:47 AM   #5
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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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