Unit fraction, equally spaced denominators not integer

I tracked down Nagell's paper in early 1992, since I had found a proof and wanted to see whether it was the same as his. (It turned out to be essentially the same idea.) Unfortunately, I've since lost my photocopy of his paper, but it came from UIUC, so that would be where I'd start looking. If I remember right, the journal where it appeared is incredibly obscure, and UIUC was the only place in North America that had a copy.

Here's a proof copied from my very old TeX file. A bit awkwardly written, but it explains how to do the case analysis, which is the part that makes this approach simpler than what Erdős did.

Suppose that $a$, $b$, and $n$ are positive integers and $$\frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+2b}+\cdots+\frac{1}{a+nb} = c \in \mathbb{Z}$$ We can take $\gcd(a,b) = 1$ and $a > 1$, and it is easy to show that $n > 2$.

Suppose that $b$ is odd. In the arithmetic progression, there is then a unique number $a + mb$ divisible by the highest possible power of $2$, because for all $k$, the progression runs through a cycle modulo $2^k$ which contains each value exactly once. Then multiplication by $\ell = \hbox{lcm}(a,a+b,\ldots,a+nb)$ gives $$\frac{\ell}{a}+\frac{\ell}{a+b}+\frac{\ell}{a+2b}+\cdots+\frac{\ell}{a+nb} = \ell c.$$ All of the terms here except $\frac{\ell}{a+mb}$ are even. Thus, $b$ cannot be odd.

Now suppose that $b$ is even, and $b \le \frac{n-2}{3}$. By Bertrand's Postulate, there is a prime $p$ such that $\frac{n+1}{2} < p < n+1$. Then $p$ does not divide $b$, and $p$ is odd. We must have $a \le n$, because $$1 \le c = \frac{1}{a}+\frac{1}{a+b}+\cdots+\frac{1}{a+nb} < \frac{n+1}{a}.$$

Since $b$ generates the additive group modulo $p$ and $n+1 > p$, at least one of the numbers $a, a+b, \dots, a+nb$ is divisible by $p$. At most two are, since $2p > n+1$. Suppose that $p$ divides only the term $a+kb$. Then $$\frac{1}{a+kb} = c-\sum_{j \neq k}{\frac{1}{a+jb}}.$$ The denominator of the left side is divisible by $p$, but that is not true of the right side. Thus, $p$ must divide two terms.

Now suppose that $p$ divides $a+{\ell}b$ and $a+kb$, with $mp = a+{\ell}b < a+kb = (m+b)p$. Then $$\frac{1}{p}\left(\frac{2m+b}{m(m+b)}\right) = c - \sum_{j \neq \ell,k}{\frac{1}{a+jb}}.$$ This implies that $p \mid (2m+b)$. However, $$a+{\ell}b \le n + (n-p)\left(\frac{n-2}{3}\right) < n + \frac{n}{2}\left(\frac{n-2}{3}\right).$$ Therefore, $$m < \frac{a+{\ell}b}{n/2} < \frac{n-2}{3} + 2.$$ It follows that $2m+b \le n+1$. However, $n+1 < 2p$, so $2m+b = p$. This contradicts the fact that $b$ is even and $p$ is odd.

Finally, suppose that $b$ is even, and $b \ge \frac{n-1}{3}$. Since $a > 1$ and $a$ is odd, we must have $a \ge 3$. We must also have $b \ge 4$, since if $b=2$, then Bertrand's Postulate and the fact that $n \ge a$ (as above) imply that one of the terms is a prime, which does not divide any other term.

Now, we show that $n \le 13$. To do that, note that $$\frac{1}{a}+\frac{1}{a+b}+\cdots+\frac{1}{a+nb} < \frac{1}{3} + \frac{1}{7} + \frac{3}{n-1}\left(\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right).$$ A simple computation shows that for $n \ge 14$, $$\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} < \frac{11}{63}(n-1),$$ which implies that the sum is less than 1. Hence, we must have $n \le 13$.

Thus, the sum is at most $$\frac{1}{3} + \frac{1}{7} + \frac{1}{11} + \cdots + \frac{1}{47} + \frac{1}{51} + \frac{1}{55} < 1.$$

Therefore, the sum is never an integer, and the theorem holds.

If you are willing to use some heavier results, the fact that $$\sum_{k=0}^{n}\frac{1}{m+kd}$$ is never an integer follows from the following theorem by Shorey and Tijdeman (which refines a theorem of Sylvester):

The greatest prime factor of the product $m(m+d)\cdots (m+nd)$ is greater than $n+1$ unless $(m,d,n)=(2,7,2).$

This is proven in "On the greatest prime factor of an arithmetical progression", and refinements are given in subsequent papers of the authors, finding these references shouldn't be that hard. It implies your result because it shows that one of the fractions has a denominator divisible by a prime $p$ which doesn't appear in any other denominator.

You can cite H Belbachir and A Khelladi, On a sum involving powers of reciprocals of an arithmetic progression, Ann Math Inform 34 (2007) 29-31, MR 2009d:11018, where a more general result is given. If you are OK with Russian, there is Z D Gorskaya, On an arithmetic property of a harmonic sum, Ukrain Mat Z 6 (1954) 375-384, MR 16, 998j.

Nagell wrote a nice intro Number Theory textbook, which was republished by the American Math Society. Maybe the result is in it.

EDIT: I have had a look at the Belbachir and Khelladi paper, at http://www.kurims.kyoto-u.ac.jp/EMIS/journals/AMI/2007/ami2007-belbachir.pdf and I find that it rests heavily on the Shorey and Tijdeman paper cited in Gjergji Zaimi's answer.

FURTHER EDIT: I think that Erdos himself proves the result in a paper of 1932 (but it's in Hungarian), Egy Kurschak-fele elemi szamelmeleti tetel altalanositasa, Lapok 39 (1932) 17-24 (my apologies for omitting the numerous diacritical marks). This is freely available, with a summary in German at the end, at http://www.renyi.hu/~p_erdos/1932-02.pdf It would seem that in 1932 Erdos was unaware of Nagell's work.