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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; the edition read is named on the source card. The paper is in Hungarian, with Russian and English summaries (pp. 39-40). It numbers no theorem; the result is the conjecture stated in its opening paragraph.

Statement

Main theorem (p. 33). Let

$$q(y)=\prod_{m=0}^{n}(y-a_m)$$

have only real zeros, equally spaced: $a_m-a_{m-1}=d$ for $m=1,2,\ldots,n$, with $d$ a constant. Then the distances between pairs of consecutive zeros of the derivative $q'(y)$ increase monotonically from the midpoint $M=(a_0+a_n)/2$ of the interval $(a_0,a_n)$ toward the endpoints of that interval.

The paper attributes the conjecture to Pál Erdős and gives no reference for it. The Russian summary (p. 39) states the same with $(a_0,a_n)$; the English summary (pp. 39-40) writes the midpoint as $(a_0+a_m)/2$ and the interval as $(a_0,a_m)$, reusing the product's index $m$.

Normalization (pp. 33-34). The affine change $y=a_0+dx$ carries the zeros $a_0,\ldots,a_n$ to $0,1,\ldots,n$, so it suffices to treat $p(x)=x(x-1)(x-2)\cdots(x-n)$. The zeros of $p'$ are the roots of

$$\frac1x+\frac1{x-1}+\cdots+\frac1{x-n}=0, \tag{3}$$

written $t_1<t_2<\cdots<t_n$ with $k-1<t_k<k$ for $k=1,\ldots,n$. In this notation the theorem says that the differences $t_{k+1}-t_k$ increase as the gap moves away from $n/2$ on either side.

Lemma 1 (1. segédtétel, p. 34). The zeros of $p'$ lie symmetrically about the midpoint $n/2$ of $(0,n)$: if $t_k$ is the root of (3) in $(k-1,k)$, then $s_k=n-t_k$ is the root of (3) in $(n-k,n-k+1)$. The paper notes that for even $n$ the point $n/2$ is a zero of $p$ and not of $p'$, and for odd $n$ it is a zero of $p'$; by the symmetry it suffices to study the roots in one half, $(n/2,n)$.

Lemma 2 (2. segédtétel, p. 35). The function $f(x)=\sum_{m=0}^{n}1/(x-m)$ is positive on $(k-1,t_k)$ and negative on $(t_k,k)$.

(III) (Az Erdős-sejtés bizonyítása, pp. 36-39). In the half $(n/2,n)$, for $k-1>n/2$,

$$t_{k+1}-t_k>t_k-t_{k-1},$$

equivalently $(t_{k+1}+t_{k-1})/2>t_k$.

The range is the print's. For odd $n$ the point $n/2$ is the zero $t_{(n+1)/2}$, and (III) with Lemma 1 compares every pair of adjacent gaps on each side of it. For even $n$ the gap $(t_{n/2},t_{n/2+1})$ contains the midpoint, and (III) as printed compares only the gaps from $(t_{n/2+1},t_{n/2+2})$ outward; the print does not compare the gap containing the midpoint with its neighbours.

Proof pointer

Lemma 1 (p. 34) follows by substituting $x=n-s$ in (3) and reversing the order of summation. Lemma 2 (p. 35) writes $f(x)=f(x)-f(t_k)$ as $(t_k-x)$ times a sum of positive terms on $(k-1,k)$. Step (I) (p. 35) gives $k-\tfrac12<t_k$ for $k-1\ge n/2$, and step (II) (p. 36) gives $t_k+1<t_{k+1}$ on $(n/2,n)$; see (I) and (II). For (III) (pp. 36-39), the midpoint $(t_{k+1}+t_{k-1})/2$ lies in $(k-\tfrac12,k)$, as $t_k$ does, so by Lemma 2 it suffices that $f_k=2f\bigl(\tfrac{t_{k+1}+t_{k-1}}2\bigr)-f(t_{k+1})-f(t_{k-1})$ is negative. The paper writes $f_k=\sum_m A_m$ with explicit terms, finds the sign of each $A_m$ (negative for $m<k-1$ and $m=k$, positive for $m=k-1$ and $m>k$), and pairs $A_{k-1-r}$ with $A_{k+r}$ for $0\le r\le n-k$; the remaining terms, with $m\le 2k-n-2$, are all negative, and each pair is shown negative using (I) and (II) and the monotonicity of $x/(x^2-a^2)$ for $x>a$.

Read depth. Claims checked: the statement, the normalization, Lemmas 1 and 2 and the range of (III) were read clause by clause on the page images of the print, and the proof was followed but not checked step by step. Nothing here is independently reviewed.

Dependencies

(I) and (II) (pp. 35-36). No outside results are cited.

Bears on

• Problem 1114: the theorem is the problem's statement for a polynomial of degree $n+1$ with zeros $a_0<\cdots<a_n$ and the interval $(a_0,a_n)$, as Erdős's conjecture stated in the paper's opening paragraph. For even $n$ the print's step (III) does not compare the gap containing the midpoint with its neighbours, as noted above.