Feels like I am probably missing something obvious.
Are there distinct primes $p,q$ and positive integers $m,n$ such that
$$ \sum_{i=0}^{n} p^i = \sum_{j=0}^{m} q^j$$
Guessing the answer is no, but unable to prove it.
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2$\begingroup$ It can certainly happen if you don't insist on starting at $p^0$ and $q^0$: $4 + 8 = 3 + 9$. $\endgroup$– LSpiceCommented Jul 30, 2023 at 18:25
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3 Answers
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Richard Guy, Unsolved Problems in Number Theory, 3rd Ed., section D10, writes,
The conjecture of Goormaghtigh, that the only solutions of $$ {x^m-1\over x-1}={y^n-1\over y-1} $$ with $x,y>1$ and $n>m>2$ are $\{x,y,m,n\}=\{5,2,3,5\}$ and $\{90,2,3,13\}$ is still open; some results have been obtained by Le Mao-Hua....
Guy cites several papers of Le Mao-Hua:
On the Diophantine equation ${x^3-1\over x-1}={y^n-1\over y-1}$, Trans Amer Math Soc, 351 (1999) 1063-1074, MR 99e:11033 [but see Leu Ming-Guang and Li Guan-Wei, The Diophantine equation 2x^2+1=3^n$, Proc Amer Math Soc, 131 (2003) 3643-3645 (electronic)].
The exceptional solutions of Goormaghtigh's equation ${x^3-1\over x-1}={y^n-1\over y-1}$, J Jishou Univ Nat Sci Ed, 22 (2001) 29-32, MR 2002d:11036.
On Goormaghtigh's equation ${x^3-1\over x-1}={y^n-1\over y-1}$, Acta Math Sinica, 45 (2001) 505-508, MR 2003f:11045.
Exceptional solutions of the exponential Diophantine equation ${x^3-1\over x-1}={y^n-1\over y-1}$, J Reine Angew Math, 543 (2002) 187-192, MR 2002k:11042.
Actually, Guy cites over 40 papers of Li Mao-Hua, some of which have titles like "An exponential Diophantine equation" and might possibly be concerned with the equation we are discussing.
The Wikipedia page on Goormaghtigh's conjecture, https://en.wikipedia.org/wiki/Goormaghtigh_conjecture contains some more references.
answered Jul 31, 2023 at 4:23
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2$\begingroup$ The question was also raised a few years back at math.stackexchange.com/questions/2599653/… and a question on this website, mathoverflow.net/questions/312105/… $\endgroup$ Commented Jul 31, 2023 at 4:27
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$\begingroup$ Also mathoverflow.net/questions/122074/goormaghtigh-equation $\endgroup$ Commented Jul 31, 2023 at 4:33
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2$\begingroup$ A more recent paper on aspects of Goormaghtigh's conjecture is, An old and new approach to Goormaghtigh’s equation by Michael A. Bennett, Adela Gherga and Dijana Kreso, Trans. Amer. Math. Soc. 373 (2020), 5707-5745, DOI: doi.org/10.1090/tran/8103, MathSciNet review: 4127890 $\endgroup$ Commented Jul 31, 2023 at 4:55
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I'm guessing that $1+5+25=31=1+2+4+8+16$ is the only example. There are certainly no more small examples, and probabilistically they get rare very quickly. But I only checked the first 80 primes to the first 80 powers. However, proving it might be quite tricky.
answered Jul 30, 2023 at 21:26
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4$\begingroup$ There is only one solution for $(n, m) = (2, 4)$. $p(p + 1) = q(q + 1)(q^2 + 1)$. If $p, q$ is odd then $p \mid q^2 + 1$. So $p \equiv 1 \pmod{4}$. And we get contradiction modulo 4. $\endgroup$ Commented Jul 30, 2023 at 21:59
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Going a bit outside the box here.
If we define a "Fibonacci-base" representation as
$$({\ldots}a_2a_1a_0)_F=\sum_{k=1}^\infty a_kF_k$$
where $a_k\in\{0,1\}$ and $F_k$ are the Fibonacci numbers, then applying the Shanks transformation to ${\ldots}111_F$ gives $-1$. Applying the Shanks transformation to the ordinary binary representation ${\ldots}111_2$ also gives $-1$.
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2$\begingroup$ TeX note: please use
\sum, not\Sigma, for sums; compare $\displaystyle\sum_{k = 1}^\infty$\sum_{k = 1}^\inftyto $\displaystyle\Sigma_{k = 1}^\infty$\Sigma_{k = 1}^\infty. \\ I'm sorry that I'm too dense to understand: how does this answer the question? $\endgroup$– LSpiceCommented Jul 31, 2023 at 19:19 -
1$\begingroup$ I see. When I tried, I used \Sum and got errors. Problem nust have been the case. $\endgroup$ Commented Jul 31, 2023 at 19:53