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On a certain Diophantine equation
编辑:林煜发布时间:2019年01月11日

Speaker:Shaofang Hong (Sichuan University)

Time:2019-01-11 10:40

Location:Conference Room 108 at Experiment Building at Haiyun Campus

Abstract: Let $S=\{s_i\}_{i=1}^{\infty}$ be any given infinite sequence of positive integers (note that all the $s_i$ are not necessarily distinct not necessarily monotonic). Let $f(x)$ be a polynomial of nonnegative integer coefficients. For any integer $n\ge 1$, one lets $S_n:=\{s_1, ..., s_n\}$ $H_f(S_n):=\sum_{k=1}^{n}\frac{1}{f(k)^{s_{k}}}$. When $f(x)=x$ $s_i=1$ for all positive integers $i\ge 1$, then Theisinger proved in 1915 that $H_f(S_n)$ is not an integer if $n\ge 2$. Equivalently, this tells us that the Diophantine equation $1+\frac{1}{2}+...+\frac{1}{n} =m$ has only the solution $(m,n)=(1,1)$. In 1923, Nagell extended Theisinger's theorem by showing that if $s_i=1$ for all positive integers $i\ge 1$ $f(x)=ax+b$ with $a\ge 1$ $b\ge 0$ being integers, then $H_f(S_n)$ is not an integer if $n\ge 2$. Hence the Diophantine equation $\sum_{i=0}^{n-1}\frac{1}{a+bi}=m$ has only the solution $(a,b,m,n)=(1,b,1,1)$. In this talk, we will report recent progress on the Diophantine equation $\sum_{k=1}^n\frac{1}{f(k)^{s_k}} =m$, where $f(x)$ is a polynomial of nonnegative integer coefficients.


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