English translation of Erdős proof of: $\sum_{k=0}^{n} \frac{1}{a+kd} $ is not an integer. In his paper, Erdős proved that: 


  For positive integers $a$ and $d$, with $\gcd(a, d)=1$, and positive integer $n$, the expression 
  $$\frac{1}{a}+\frac{1}{a+d}+\dots+\frac{1}{a+nd}$$
  is not an integer.


Do you know a reference for an English language of this proof?
I can't understand Hungarian but I have an idea of what it's talking about. 

This is a proof by contradiction. Suppose that for some $n$, the fractional sum posted in the OP is an integer.

Let $p &gt; n$ be a prime such that $p^{\alpha} || a + kd$ (Erdos will prove that such a prime exists later in the paper). It's obvious that $(p, d) = 1$ because if not, then $(a, d) \neq 1$. 

He starts the proof making the statement:

Claim:
There cannot exist $k'$ such that $k \neq k'$, $k' &lt; n$ and $p^{\alpha} | a + k'd$

Proof:
Suppose that such a $k'$ exists. Then by simple modular arithmetic,
$$k \equiv k' \pmod{p^\alpha}$$ This proves the claim since $|k - k'| &lt; n &lt; p^\alpha$ is absurd.
$\square$ 
$$\\$$
Denote 
$$\prod_{i = 0}^n (a + id) = !(a + nd)$$

The fractional sum (let's call it $S$) is thus equal to 
$$S = \frac{!(a + nd) + \frac{a \cdot !(a + nd)}{a + d} + \cdots + \frac{a \cdot !(a + nd)}{a + nd}}{a \cdot !(a + nd)}$$

It's clear that $v_p\left(\frac{a \cdot !(a + nd)}{a + kd}\right) &lt; v_p\left(\frac{a \cdot !(a + nd)}{a + ld}\right)$ for $l \neq k$ so $v_p(S) = v_p\left(\frac{a \cdot !(a + nd)}{a + kd}\right) &lt; v_p(a \cdot !(a + nd))$ so it's clear that $S$ cannot be an integer. This is a contradiction.

Next is the step that takes up most of the proof: the proof of the prime $p &gt; n$. Note that $p$ does not necessarily have to be greater than $n$; only $p^\alpha$ has to necessarily be greater than $n$. 

Suppose that for all $a + id$ that if a prime $p$ divides it, then the maximal power of $p$ that divides it is less than or equal to $n$. Now consider the fraction
$$\frac{!(a + nd)}{n!}$$
If $q$ is a prime such that $\sqrt{n} &lt; q \le n$, then at most $\lfloor \frac{n}{q} \rfloor$ numbers less than or equal to $n$ are divisible by $q$. Since $q &gt; \sqrt{n}$, each of those numbers can be divisible by $q$ at most once. Hence, there are $\lfloor \frac{n}{q} \rfloor$ factors of $q$ in $n!$. Since $q^2 &gt; n$, under the assumption of the statement we are trying to prove false, there are at most $\lfloor \frac{n}{q} + 1 \rfloor$ (Exercise for the reader: why $ + 1$?) factors of $q$ in $!(a + nd)$. Using similar logic, Erdos deduces that 

$$\frac{!(a + nd)}{n!} \le \prod_{\sqrt{n} &lt; q \le n} q \prod_{n^{1/3} &lt; q \le \sqrt{n}} q^2 \prod_{n^{1/4} &lt; q \le n^{1/3}} q^3 \cdots = \prod_{q \le n} q \prod_{q \le \sqrt{n}} q \prod_{q \le n^{1/3}} q \cdots$$

In addition, we have that 
$$\frac{!(a + nd)}{n!} &gt; d^n \ge 4^n$$ 
assuming that $d \ge 4$ (Exercise for the reader: prove the original statement by the OP for $d &lt; 4$). 

Hence,

$$4^n &lt; \prod_{q \le n} q \prod_{q \le \sqrt{n}} q \prod_{q \le n^{1/3}} q \cdots$$

The rest of the proof is devoted to showing that such an inequality is impossible. Let 
$$a_i = \lceil \frac{n}{2^i} \rceil$$
where $a_i \ge 1$. We can easily prove that $a_i \le 2a_{i + 1}$. Hence, the intervals, $(a_i^{1/\alpha}, (2a_i)^{1/\alpha})$ for all (valid) $i$ covers the interval $(1, n^{1/\alpha})$. Consider the binomial: ${2n \choose n}$. Any prime number $n^{1/\alpha} &lt; p \le (2n)^{1/\alpha}$ (\alpha \ge 1) must divide the binomial at least once because $\lfloor \frac{2n}{p} \rfloor &gt; 2 \lfloor \frac{n}{p} \rfloor$ (and also by Legendre's formula). 

Thus, 
$$\prod_{q \le n} q \prod_{q \le \sqrt{n}} q \prod_{q \le n^{1/3}} q \cdots &lt; \prod {2a_i \choose a_i}$$

We will show that $\prod {2a_i \choose a_i} &lt; 4^n$. This is easy since by simple induction (or by Stirling's formula if you want to drop the sledgehammer), we have that ${2n \choose n} &lt; 4^{n - 1}$ if $n \ge 5$. Summing up all of the powers of $4$ in the RHS of
$$\prod {2a_i \choose a_i} &lt; \prod 4^{a_i - 1}$$ 
we have that 
$$\prod 4^{a_i - 1} &lt; 4^n$$
(Proof is left as an exercise to the reader). Hence, 
$$\prod {2a_i \choose a_i} &lt; 4^n$$
This is a contradiction to an earlier inequality. So there must exist a prime power $p^\alpha | a + kd$ such that $p^\alpha &gt; n$. 

QED
$$\\$$

The rest of the paper is Erdos dealing with $d = 1, 2, 3$.
Here is what Google translate produces:

Page 1
Budapest, 1932. Sonderabdruck aus «Mathematical and Physical Tabs Band XXXIX. Budapest 1932. AN EXCELLENT CHURCH-EARTH BETWEEN DIALOGUE Generalizations. THEISINGER proved that the harmonic series of tranches m = n . 1 Sums, m , Can not be integers.1 OBLÁΤY general m = 2 mn It has re-established this item, 2 if it has proved that am m = k ni Can not be an integer if the ams are positive integers and (am, m) = 1 Kürschák 3 gave it a completely complete proof of it m = n 1 . The batch of that m Can not be an integer, no matter m = k The integer integer is ak and n. . This item will be generalized if k, k + 1,. . .K + n Instead of a special arithmetic sequence instead of the more general arithmetic lines, Show them a batch Let a, d, be any positive integers; the Expression can not be an integer. The bizonyításnál assume that d and relative prime numbers, Because if they are not the reciprocal of their greatest common divisor (1), the other factor is again (1), but a and d They are now relative primes. Thus, if a line (a, d) = 1, it is true, Otherwise it is valid. 1 T Heisinger Monatshefte fϋr Math. u. Ρhys., 26, k. (1915) 135. 1. 2. Oblath: Mathematical and Physical Journal, 27 k. (1918) 93. 1. 3 KURSCHAT Mathematical and Physical Journal, 27 k. (1918) 299. 1. 1
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h PALE FOREST. First, d> 4; The other cases at the end of our We'll take care of it separately. Evidence is based on the following auxiliary: the One of the numbers can be divided by a p prime number that is Greater than n. If the auxiliary evidence has been proved, further evidence Nete ​​the next. Let kd + pa is divisible, where p '> n; They can not exist in a variety of k from k ' is not possible because n. Let's introduce it now nominations. You can then write (1) The counter members are integers; As well as The tag can be divided by the lower power of p, As the other members of the counter and the denominator, so (4) can not be integer. Q. ed That is the only proof we need to prove. Contrary to this entry, suppose a + d, a + 2d, ... a + nd
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GENERATION OF A BACKGROUND OF AN EXTRAORDINARY BALANCE ELEMENTS. Series members. and only p ' n, up Under theorem every element that only up to - first Power, even up to [~] + 1 times nyezőként. After the acronym, no more than p remains. Let 1in  n, under the terms of each element that No more than a second power, even in total Up to [~] +1, up to [~] + 1 times, Such as p2, so after the abbreviation, the maximum p2 remains in the counter. In the same way it will be appreciated that if "~ + n  4
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ERDŐS PAL. We will prove that this is not possible. To this end, Let's play it Binomial coefficients. (9) The register is divided into every n  2 [k]) p p Let us denote the first integer with> xa {x} Where the arbitrary integer; then furthermore So since k and ak + 1 are integers,
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GENERATION OF A BACKGROUND OF AN EXTRAORDINARY BALANCE ELEMENTS. Be already m the first number for which it is 2m &lt;1; then m = 1. Clearly, 2a1> n. Furthermore, (10) it is Am  10 s suppose that the theorem is true min- For less than n; at Which is achieved by applying the batch to 2a2-1; 2a2 - from one item to your true since n> 2a2 - 1
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PALE FOREST. With full induction, we can easily assure that n> 5 1 ~ n) &lt; 4n-1, so (13) by Now, it is obvious that 2a1  k + 2a,
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Ι Ι Ι Ι Ι GENERATION OF A BACKGROUND OF AN EXTRAORDINARY BALANCE ELEMENTS. Our contradictions are minimal. Further proof is so Occurs as in the general case. Finally, let d = 2; Since (1) is the first member the And the rest Smaller than this, so that (i) integers can be most one of the members must be below + to stand, ie that n> the be. So it is The last member of a series of numbers 2n +> = a + 2a and 3a of the series- There are all odd numbers for which  a and 3a occur among the members (15); but Then another member of the series can not be divided Let be 3a, because 3a + 2.3a = 3a + 1 would also occur (15) - Which is impossible because of the maximality of 3a. The further Conclusion is made as in the general case. Similarly, but with slightly longer calculations, We will also consider the following item for the special cases discussed later: The arithmetic line there is a + kd member, a prime number p It contains more power than the other members. It can also be proven for special cases, ter- Of course, with the exception of d = 1, it is proven to be a general case Our auxiliary. Generally, the OBLÁTH theorem is that t. i. can not be an integer, if (k, a + kd) = 1, the method of his demonstrated. Pál Erdős.
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q Ρ. ERDŐS. VERALLGEMEINERUNG EINES ELEMENTAR-ZAHLENTHEORETISCHEN SATZES VON KIRCSCHAKK. 1 mn Herr THEISINGER bewies, dass ~ -für keinen Wert von n eine Ganze Zahl darstellt. Herr OBLQTH gab eine Verallgemeinerung dieses m = n Satzes, Inde bewies er, dass ~ am Keine ganze Zahl sein kann, 17 m = n i wenn (am, m) = 1 ist. KÜRSCHÁK bewies elementar, dass ~ -bei m = k Keinem Werte von k und n eine ganze Zahl sein kann. Hier wird dieser Satz folgendermassen verallgemeinert, indem statt Der speziellen arithmetischen Reihe Allgemeinere arithmetische Reihen betrachtet werden. E seien a, d, n beliebige positive ganze Zahlen, dann ist Keine ganze Zahl. Besteht der Grundgedanke des Beweises Darin, dass ein glied a + kd Angegeben wird, welches durch eine höhere Potenz einer Primzahl Teilbar ist, als die übrigen Glieder. Dies ergibt sich aus der Analyzes der Primteiler der Ausdrücke:. Paul Erdős.