- How to solve $3^x5^y-2^s7^t=1$ completely?
- Does there exists any general techniques dealing with such exponential equations?
- For equations like $a^x-b^y=1$, Mihăilescu's theorem (Catalan conjecture) guarantees the only solution is $3^2-2^3=1$. For equations like $\prod p_i^{n_i}-\prod q_j^{m_j}=1$, the abc conjecture asserts there are only finitely many solutions such $\prod p_i^{n_i}>(\prod p_i\prod q_j)^{1+\epsilon}$, thus only solutions with $n_i$ not very big happens often. Although the abc conjecture is not yet fully verified, this equation is far from covering all of the abc conjecture. So is there any general conclusions on such equations?
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Will Jagy
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Anonymous Coward
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You must mean $\prod_i p_i^{n_i} - \prod_j q_j^{m_j} = 1$ (with exponents $n_i$ and $m_j$, not $n^i$ and $m^j$). – Noam D. Elkies Oct 23 '13 at 16:08
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@NoamD.Elkies, thanks for pointing out that! – Anonymous Coward Oct 24 '13 at 02:34
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The only solutions of $3^x 5^y - 2^s 7^t = 1$ in nonnegative integers are:
x y s t
1 0 1 0 3 - 2 = 1
0 1 2 0 5 - 4 = 1
2 0 3 0 9 - 8 = 1
1 1 1 1 15 - 14 = 1
2 2 5 1 225 - 224 = 1
There is a large literature on such Diophantine questions. One key phrase is "$S$-unit equations". In general it has been known for some time that there are finitely many solutions, and indeed for equations of the form $$\prod_i p_i^{n_i} - \prod_j q_j^{m_j} = 1$$ this already follows from Thue's theorem (1909); and by now we even have effective algorithms known to find all solutions. There's still no elementary technique known in general, but in your case (where only the primes 2,3,5,7 appear) David Rusin reports that an elementary solution is contained in a 1976 paper
L. J. Alex: Diophantine equations related to finite groups, Communications in Algebra 4 #1 (1976), 77-100 (MR54:12634).
Noam D. Elkies
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Excuse me, but how do we fit this equation into Thue's theorem? – Anonymous Coward Oct 26 '13 at 01:01
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1An old trick: there's only finitely many possibilities for $x,y,s,t \bmod 3$; for each one we get an equation of the form $AX^3-BY^3=1$ for some nonzero constants $A,B$, and this equation has finitely many solutions by Thue even before we require that $X$ and $Y$ have no prime factors outside ${2,7}$ and ${3,5}$ respectively. – Noam D. Elkies Oct 26 '13 at 03:04