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Statement

Setting (p. 2). Fix a positive integer $j$ and positive integers $k_1,\dots,k_j$, and consider equation (2.1),

$$\prod_{i=1}^{j}\prod_{l=0}^{k_i-1}(x_i+l)=y^2,$$

in positive integers $x_1,\dots,x_j$ subject to condition (2.2): $x_s<x_t$ implies $x_s+k_s\le x_t$. So the $i$th block is the $k_i$ consecutive integers $x_i,\dots,x_i+k_i-1$, and (2.2) keeps blocks with distinct starting points disjoint. The paper assumes without loss of generality that $j>1$ (the case $j=1$ is the Erdős--Selfridge theorem) and that $2\le k_1\le k_2\le\dots\le k_j$ (when $k_1=1$ the equation has infinitely many solutions trivially).

Theorem 2.1 (p. 2). "If either $k_1 = 2$ or $(k_1,k_2) = (3,3)$ then equation (2.1) has infinitely many solutions with (2.2)."

The paper calls this a generalization of Theorem 1 of Ulas (Enseign. Math. 51 (2005), 331--334), and notes that it covers cases Erdős and Graham left out of their question, whose blocks have at least four integers each.

Proof pointer

Section 3, pp. 3--4. The blocks other than the first one (when $k_1=2$), or other than the first two (when $(k_1,k_2)=(3,3)$), are fixed so that their product is a squarefree number times a square, in general a factorial; the remaining one or two blocks then reduce the equation to a Pell equation with infinitely many solutions. For $j=2$ and $(k_1,k_2)=(3,3)$ the paper uses an observation of K. R. S. Sastry recorded in Guy's Unsolved Problems in Number Theory.

Read depth

Claims checked: the setting, the hypotheses and the statement were read clause by clause on the page images of the print, and the proof was followed for structure. Nothing here is independently reviewed.

Dependencies

None in the corpus. The proof uses the solvability of Pell equations and Bertrand's postulate.

Source. M. Bauer and M. A. Bennett, On a question of Erdős and Graham, Enseign. Math. (2) 53 (2007), 259--264; the edition read, paged 1--6, is named on the source card.

Bears on

• Problem 363: the problem asks about blocks of at least four integers each. Theorem 2.1 treats shortest block lengths two and three, which the problem excludes, so it is not a counterexample to the problem.