Kürschâk and Nagel’s theorems (Erdos 1932)

\displaystyle \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+ \frac{1}{n-1}+\frac{1}{n}

\displaystyle \frac{1}{m+1}+\frac{1}{m+2}+\frac{1}{m+3}+\cdots+ \frac{1}{m+n-1}+\frac{1}{m+n}

\displaystyle \frac{1}{m+k}+\frac{1}{m+2k}+\frac{1}{m+3k}+\cdots+ \frac{1}{m+(n-1)k}+\frac{1}{m+nk}

\displaystyle \frac{a_1}{m+1}+\frac{a_2}{m+2}+\frac{a_3}{m+3}+\cdots+ \frac{a_{n-1}}{m+n-1}+\frac{a_n}{m+n}, ~~(a_i, m+i)=1

None of the above quantities are integers.

Proof: For the first expression, look at the largest prime- when we clear denominators, the denominator is divisible by this prime and numerator is not.
For the second expression, if the n is smaller then m then the quantity is less than one, otherwise there will be a prime between $m$ and $2m$ (Some weak versions of Prime Number Theorem like Betrand Postulate is enough to see that)- and now again taking common denominators, we see that denominator is divisible by largest prime in that interval and the numerator is not.

Same argument even for the fourth expression.

Another way: Idea: Look at the highest power of two that divides one of the denominators. Observation: There is a unique number in that list which is divisible that power of 2- for if there are two, there is a number between them divisible by a higher power of 2. And now carry out the argument with this power of 2 instead of the largest prime- look at the numerator and denominator modulo that power of 2.

For the third expression- Even here we adopt the previous argument and try to find prime with more valuation for denominator than the numerator-
We need some information about some number of primes- in intervals, arithmetic progressions etc- Elementary arguments involve understanding factorial like expressions (similar to Tchebyshev’s ideas for prime counting) \displaystyle \frac{(a+d)(a+2 d) \ldots(a+n d)}{n !}

Click to access 1932-02.pdf

https://users.renyi.hu/~p_erdos/1932-02.pdf

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