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Statement

Equation (1.2), its normalization $0\le\alpha<l$ and its trivial solutions are as on Theorem 3 (p. 374).

Theorem 4 (p. 375). Let $2\le k\le5$ and $l\ge3$. The only non-trivial solutions of (1.2) with $\alpha=0$ are given by $k=l=3$ and

$$(x,z)\in\{(-2/3,2/3),\ (-4/3,2/3)\}.$$

The paper notes (p. 375) that Sander (J. London Math. Soc. 59 (1999), 422--434) proved that (1.2) with $\alpha=0$ has no solution for $k=2,3,4$; the two solutions for $k=l=3$ are missing from his Proposition 2, so his Conjecture 1, that for $k\ge3$ equation (1.2) with $\alpha=0$ has only the trivial solutions, should be modified. The case $k=5$ is new.

Source. K. Győry, L. Hajdu and N. Saradha, On the Diophantine equation $n(n+d)\cdots(n+(k-1)d)=by^l$, Canad. Math. Bull. 47 (2004), no. 3, 373--388, doi:10.4153/CMB-2004-037-1; Theorem 4 on p. 375, its proof on p. 384. The edition is recorded on the source card.

Read depth. Claims checked: the statement was read clause by clause against the published print, and the proof on p. 384 for its structure only. A second reader checked the statement, hypotheses, label and page against the print.

Proof pointer

Section 5, p. 384. With $\alpha=0$, (1.3) holds with $\beta=\gamma=0$, and Theorem 3 leaves only $k=l=3,4,5$ and $k=2$, $l=4$. Theorem 8(i) gives $(n,d)\in\{(-4,3),(-2,3)\}$ for $k=l=3$, hence the two solutions; Theorem 9(ii) and Theorem 8(iii) exclude $k=l=4$ and $k=l=5$; Lemma 7 (p. 379) excludes $k=2$, $l=4$. Bennett, Bruin, Győry and Hajdu (Proc. London Math. Soc. (3) 92 (2006), p. 292) say the proofs of Theorems 8 and 9 for $l=3$ depend on an incorrect lemma of this paper (Lemma 6); the case $k=l=3$ here uses Theorem 8(i) with $l=3$.

Dependencies

Theorem 3, Theorem 8 parts (i) and (iii), Theorem 9(ii) and Lemma 7 of the same paper.

Bears on

No problem page of this corpus.