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 September 20th, 2011, 05:38 AM #1 Newbie   Joined: Sep 2011 Posts: 18 Thanks: 0 Exponents - conditions Hi! As I was solving a math problem (an equation with exponents) I found myself unable to go any further without having determined what conditions must be met by integers $m,n,x$ so that $5^m-5^n=2^x$. I can post the original quetion, if needed. Thanks, Arnold
 September 20th, 2011, 06:01 AM #2 Member   Joined: Apr 2010 Posts: 65 Thanks: 0 Re: Exponents - conditions $5^1-5^0=2^2$ $for\quad k\ge 1$ $5^{2k+1}-5^{2k}=(5j)^2$
September 20th, 2011, 06:48 AM   #3
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Re: Exponents - conditions

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 Originally Posted by arnold Hi! As I was solving a math problem (an equation with exponents) I found myself unable to go any further without having determined what conditions must be met by integers $m,n,x$ so that $5^m-5^n=2^x$.
m = 1, n = 0.

It's clear that m > n, else the left side will not be positive. Further, n = 0 else the left side will be a multiple of 5. So the problem becomes
$5^m=2^x+1.$

But then you have two consecutive powers which can only happen if a base or exponent is 0 or 1, or else if the numbers are 8 and 9, by Catalan's conjecture (now a theorem).

Cannon, meet fly.

 September 20th, 2011, 08:10 AM #4 Newbie   Joined: Sep 2011 Posts: 18 Thanks: 0 Re: Exponents - conditions Thanks a lot!
 September 20th, 2011, 08:38 AM #5 Newbie   Joined: Sep 2011 Posts: 18 Thanks: 0 Re: Exponents - conditions Could you explain it to me why n=0? How do I prove it?
 September 20th, 2011, 08:46 AM #6 Newbie   Joined: Sep 2011 Posts: 18 Thanks: 0 Re: Exponents - conditions I mean, I understand why n=0. I just want to know if it can be proven somehow.
September 20th, 2011, 08:56 AM   #7
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Re: Exponents - conditions

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 Originally Posted by arnold I mean, I understand why n=0. I just want to know if it can be proven somehow.
If n > 0, then either the left is negative or a multiple of 5. Powers of 2 are never negative and are never multiples of 5.

 September 20th, 2011, 08:59 AM #8 Newbie   Joined: Sep 2011 Posts: 18 Thanks: 0 Re: Exponents - conditions Ok, thanks. So I don't need to apply modulo, etc, do I?
September 20th, 2011, 10:40 AM   #9
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Re: Exponents - conditions

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 Originally Posted by arnold Ok, thanks. So I don't need to apply modulo, etc, do I?
Depending on how you formalize it you might use the modulus operator to express "divisible by 5".

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