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Tong Lingling's user avatar
Tong Lingling
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Where did this theorem appear?
Thank you :) perhaps this theorem on $S$-unit equation is a bit stronger than the above corollary of Pólya's theorem, since the former probably could imply that $P(q_{k+1}-q_k)\to\infty$, not only $q_{k+1}-q_k\to\infty$.
awarded
 Popular Question
comment
Where did this theorem appear?
@quarague it refers to the latter. The former means $\displaystyle\limsup_{n\to\infty}q_{n+1}-q_{n} = +\infty$, instead of $\displaystyle\lim_{n\to\infty}q_{n+1}-q_{n} = +\infty$. It's easier to prove the former.
awarded
 Nice Question
comment
Where did this theorem appear?
oops! ... sorry, the link related to this answer should be ams.org/journals/mcom/1999-68-225/S0025-5718-99-01024-8/…
comment
Where did this theorem appear?
Many thanks for your answer, and for Desmond Weisenberg's research. I'll forward this result to the relevant MSE post.
accepted
Where did this theorem appear?
asked
Where did this theorem appear?
comment
A limit related with sum of digits
The solution seems not complex but I have not noticed it at all. Thanks for your amazing answer !
accepted
A limit related with sum of digits
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Prime products and powers
I'm glad it helps :)
revised
Prime products and powers
improved formatting
revised
A limit related with sum of digits
improved formatting
comment
Is there a Carmichael number $n$ satisfying $\sigma(n)\equiv0 \pmod{n+1}$?
@LSpice Thanks for your suggestions and editing.~ I have added \displaystyle before \prod and \sum. I should get more familiar with LaTeX.
revised
Is there a Carmichael number $n$ satisfying $\sigma(n)\equiv0 \pmod{n+1}$?
improved formatting
revised
A limit related with sum of digits
Supplement the proof of the bounds for the average value of $F_b(m)$
revised
A limit related with sum of digits
Supplement the proof of the bounds for the average value of $F_b(m)$
revised
Is there a Carmichael number $n$ satisfying $\sigma(n)\equiv0 \pmod{n+1}$?
Add more details about such $n$
answered
Prime products and powers
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Prime products and powers
If there exist a prime $q<p_p$ such that Legendre symbol $(\frac{p}{q})=1$ while $(\frac{-1}{q})=-1$, then it is impossible that $p^{k-1}\equiv-1\text{(mod}$ $q)$, which could be a contradiction for equation $\mathbf P_p = p+p^k$.
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