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Statement

Equation (1.1) and its hypotheses are as on Theorem 1: positive integers $n,d,y,b$, integers $l,k\ge2$, $\gcd(n,d)=1$, $P(b)\le k$ and $b$ free of $l$th powers, with $\Pi=n(n+d)\cdots(n+(k-1)d)$ (p. 373). Write $2^a\parallel\Pi$ when $2^a$ divides $\Pi$ and $2^{a+1}$ does not.

Theorem 2 (p. 374).

• (i) Let $k=4$. If (1.1) holds with $l\ge3$ and $P(b)\le2$, then $l$ has a prime factor greater than $3$ and $8\parallel\Pi$.
• (ii) Let $k=5$. If (1.1) holds with $l\ge3$ and $P(b)\le2$, then $l$ has a prime factor greater than $3$, and either $8\parallel\Pi$ or $16\parallel\Pi$.

The paper introduces it (p. 374) as showing more than Theorem 1 for $l\ge3$.

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 2 on p. 374, 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. For $k=4$, Theorem 8(ii) gives $8\parallel\Pi$ whenever $l$ has a prime factor greater than $3$, and Theorem 9 rules out $l=3,4$ with $P(b)\le2$. For $k=5$, Theorem 8(iii) for $l\ge5$ and Theorem 9 for $l=3,4$. The case $l=3$ goes through Theorem 9(i), whose proof uses Lemma 6 (pp. 378, 382); Bennett, Bruin, Győry and Hajdu (Proc. London Math. Soc. (3) 92 (2006), p. 292) say the proofs of Theorems 8 and 9 depend on that lemma, which they call incorrect, and correct the case $l=3$ in their Section 5.

Dependencies

Theorems 8 and 9 (p. 376) of the same paper.

Bears on

No problem page of this corpus. The case $b=1$ that Problem 672 concerns is settled for $k=4,5$ by Theorem 1.