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Statement

The paper's equation (1.2) (p. 374) is

$$x(x+1)\cdots(x+k-1)=\pm2^\alpha z^l$$

in rational numbers $x$ and $z\ge0$ and integers $k\ge2$, $l\ge2$ and $\alpha$ with $-l<\alpha<l$. The paper restricts to $0\le\alpha<l$ (p. 374), reducing a negative $\alpha$ to that range by a change of $\alpha$ and of $z$ to $z/2$. The solutions $x=-j$, $z=0$ with $0\le j<k$, which occur for each $\alpha$, are called trivial.

Theorem 3 (p. 375). Let $2\le k\le18$ and $l\ge3$ with $\gcd(l,k)=1$. If (1.2) holds with $z\ne0$, then $k=2$ and

$$(x,z,\alpha)\in\{(-1/2,1/2,l-2),\ (-2,1,1),\ (1,1,1)\}.$$

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; equation (1.2) on p. 374, Theorem 3 on p. 375, its proof on p. 384. The edition is recorded on the source card.

Read depth. Claims checked: the statement and the setting of (1.2) were 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. Writing $x=n/d$ and $z=y/y_1$ in lowest terms turns (1.2) into the integral system (1.3) of p. 374, $n(n+d)\cdots(n+(k-1)d)=\pm2^\beta u^l$ with $v^l=2^\gamma d^k$ and $\beta+\gamma=\alpha$, an instance of (1.1) with $P(b)\le2$. Since $\gcd(l,k)=1$, the second equation gives $d=2^hd_1^l$ with $h\ge0$, and Theorem 10 (p. 377) then yields the listed solutions.

Dependencies

Theorem 10 (p. 377) of the same paper, through Lemmas 2, 3 and 8 (pp. 377--379).

Bears on

No problem page of this corpus.