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I have solved the following three problems of my book:

Prove that the $n$th harmonic number $H_n = \sum\limits^n_{k=1} \dfrac{1}{k}$ is never an integer for $n \ge 2.$

Prove that none of the $2^{n+1}$ numbers $\pm \frac{1}{k} \pm \frac{1}{k+1} \pm \cdots \pm \frac{1}{k+n}$, where we consider all combinations of plus and minus signs, is an integer, for any positive integers $n$ and $k$.

Prove that $\sum\limits^n_{k=1} \dfrac{1}{2k-1}$ is never an integer for $n>1.$

These problems can be solved by at least two different ways: One, for example on the first problem, considering the highest power of 2 on the set $\{1, 2, \ldots, n\}$ and then checking divisibility by 2 supposing it is an integers, and the second method is to prove by induction that there is integers $j, \ell$ such that $H_n = \frac{2j+1}{2\ell}.$ With these ideas I've got this generalization:

If $p$ is a prime number, then $\sum\limits^n_{k=1} \dfrac{1}{(p-1)k - (p-2)}$ is never an integer for $n>1.$

Then I considered this more generic problem:

Problem. Show that $\sum\limits^n_{j=1} \dfrac{1}{k + mj}$ is never an integer for positive integers $n, m, k.$

But on this problem, the two later ideas doesn't work in general. I've found in "mathworld.wolfram" website that Erdős proved it on 193X or something. How?

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  • $\begingroup$ Probably you are looking for this paper: renyi.hu/~p_erdos/1932-02.pdf It is in Hungarian, so you might not understand it. The main lemma is that one of the denominators is divisible by a $p^{\alpha}>n$. I will go to sleep now, if no one will do it, tomorrow i can translate a sketch of the proof. $\endgroup$ Dec 18, 2013 at 23:32
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    $\begingroup$ mathoverflow.net/questions/39326/… $\endgroup$ Dec 18, 2013 at 23:36
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    $\begingroup$ I've just supplied a proof as an answer to the question Gjergji linked to. It's essentially the same as Nagell's proof, and different from the one Erdős gave. $\endgroup$
    – Henry Cohn
    Dec 19, 2013 at 2:11
  • $\begingroup$ Thanks! Especially Henry Cohn for the solution, amazing. I just got stuck by why you could suppose $a>1$ (my bad) and later I didn't thought you meant that $a+\ell b$ and $a + k b$ were the only numbers divisible by $p$ and it was necessary for me to imply that $p \mid 2m+b$, but it wasn't too difficult to show. $\endgroup$
    – Lucas
    Dec 19, 2013 at 6:45
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    $\begingroup$ You write "Prove that any of the $2^{n+1}$ numbers ... are integers". You mean none of those $2^{n+1}$ numbers are integers, right? Generally the word "any" should be avoided, since sometimes it means "all" and sometimes it means "some". This is the first time I've seen it mean "none". :) $\endgroup$
    – KConrad
    Dec 19, 2013 at 14:34

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