My Math Forum Positive Integer Exponent Pair Puzzle

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 March 13th, 2009, 10:28 AM #1 Member   Joined: Nov 2006 Posts: 54 Thanks: 0 Positive Integer Exponent Pair Puzzle Problem Determine all possible pair(s) (p, q) positive integers that satisfy this equation: 2^p – 5^q = 3 My unsuccessful attempt is furnished hereunder as follows: p =1 yields then 5^q =-1, which is a contradiction. p=2 yields 5^q = 1, so that q = 0., a ontradiction. If p>=3, then it follows that: 3^q = 3 (Mod 8), so that q must be odd If q =1, 2^p = 8, giving p = 3 Recalling that q is odd whenever q>=3, we substitute q = 2s+ 1, and observe that:: 2^p - 5^(2s+1) = 3 Or, 2^p = 3 (Mod 5) or, p= 4t+3, where t is a non negative integer. So we have: 2^(4t+3) – 5^(2s+1) = 3 Or, 8*(16^t) - 5*(25^s) = 3 For t =1, we obtain s =1, so that: (p, q) = (7, 3) Accordingly, till now, we have : (p,q) = (7, 3); (3,1) as valid solutions to the given problem. (Of course, if p and q were nonnegative integers rather than positive integers, then we would have obtained the additional pair (2,0) ) **** From this point onwards, my attempts to proceed any further has proved futile, and consequently, I am looking for a methodology giving any further valid solution(s) or any procedure conclusively proving that no further solutions can exist for the given problem.
 March 16th, 2009, 01:46 AM #2 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 Re: Positive Integer Exponent Pair Puzzle I don't see how to prove it, but you should not look for additional solutions: If we believe the ABC-conjecture http://en.wikipedia.org/wiki/Abc_conjecture in the form "$q$ is always smaller than 2" (the biggest found up to date is 1,6...) then this gives that there can't be any solution with $t\geq 2$.
 March 17th, 2009, 06:21 AM #3 Senior Member   Joined: Dec 2008 Posts: 160 Thanks: 0 Re: Positive Integer Exponent Pair Puzzle Problem looks like modified Catalan's conjecture: $x^u - y^v= 1$ that has only solutions: $x^u= 3^2, y^v = 2^3$. Analytical proof of it was done just in 2002 by Mihailescu and involved theory of cyclotomic fields. Unless given problem is so special that has elementary solution, you, probably, should look at Mihailescu proof first.
 March 17th, 2009, 08:21 AM #4 Member   Joined: Nov 2006 Posts: 54 Thanks: 0 Re: Positive Integer Exponent Pair Puzzle Ty Peter and zolden, for your invaluable assistance and for your considered views in the foregoing matter. This problem was inspired by dansmath problem #253 (Powers Five Apart), in the following location. http://www.dansmath.com/probofwk/probar ... chor377644 The premise of the other problem possessing an elementary solution, led me to surmise a corresponding existence of an elementary solution in the current instance. As I mentioned in the original post that all my endeavours in solving the current exercise in terms of an elementary methodology failed to reach fruition. Having regard to your views, the existence of an elementary solution in the current instance does seem highly unlikely.
 March 18th, 2009, 08:56 AM #5 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Positive Integer Exponent Pair Puzzle I understand that Mihailescu's proof uses the special nature of 1 deeply, so that the proof cannot (easily?) be generalized with greater numbers.
 March 18th, 2009, 11:25 AM #6 Senior Member   Joined: Dec 2008 Posts: 160 Thanks: 0 Re: Positive Integer Exponent Pair Puzzle As of my knowledge it is a case of Pillai's conjecture that concerns a difference of perfect powers. Michailescu's proof, most probably, cannot be extended to the general case. I mentioned it just to outline the level of complexity given problem has.
 March 18th, 2009, 04:13 PM #7 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Positive Integer Exponent Pair Puzzle Agree and agree. I was showing that it seems to be even harder than that.

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