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Source. E. Bálint, Erdős Pál egy sejtésének bizonyítása [Proof of a conjecture of P. Erdős], Mat. Lapok 11 (1960), 33--40, steps (I) (p. 35) and (II) (p. 36); the edition read is named on the source card.

Statement

Setting (pp. 33-34). $p(x)=x(x-1)\cdots(x-n)$, and $t_1<\cdots<t_n$ are the zeros of $p'$, the roots of $\sum_{m=0}^{n}1/(x-m)=0$, with $k-1<t_k<k$.

(I) (p. 35). In the interval $(n/2,n)$,

$$k-\tfrac12<t_k ;$$

the proof uses the hypothesis $k-1\ge n/2$.

(II) (p. 36). In the interval $(n/2,n)$,

$$t_k+1<t_{k+1}.$$

The print states (II) for the half $(n/2,n)$; its proof uses $k-1<t_k$, so that $t_k+1$ lies in $(k,k+1)$ with $t_{k+1}$, and needs $k+1\le n$. By the symmetry of the zeros about $n/2$ (Lemma 1, p. 34) the same bounds hold in mirror form on $(0,n/2)$.

Proof pointer

(I): the paper evaluates $f(x)=\sum_m1/(x-m)$ at $k-\tfrac12$, where it is a signed sum of the reciprocals $2/(2j-1)$, and the hypothesis $k-1\ge n/2$ leaves more positive than negative terms, so $f(k-\tfrac12)>0$; Lemma 2 (p. 35) then places $t_k$ to the right of $k-\tfrac12$. (II): from $0=f(t_k)-f(t_{k+1})$ the paper derives that $t_{k+1}-t_k-1$ times a sum of positive terms equals $1/t_{k+1}+1/(n-t_k)>0$.

Read depth. Claims checked: both statements and their hypotheses were read on the page images of the print, and the proofs were followed.

Dependencies

Lemma 2 (p. 35), stated on the main theorem page.

Bears on

• Problem 1114: steps toward the main theorem, used in step (III) to sign the paired terms. (II) says each gap $t_{k+1}-t_k$ on the right half exceeds $1$; on its own it does not give the monotonicity the problem asks for.