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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

Equation (1.1) and its hypotheses are as on Theorem 1: positive integers n,d,y,bn,d,y,b, integers l,k≥2l,k\ge2, gcd⁡(n,d)=1\gcd(n,d)=1, P(b)≤kP(b)\le k and bb free of llth powers (p. 373).

Theorem 6 (p. 375), quoted: "For fixed k≥3k\geq3 and l≥2l\geq2 with k+l>6k+l>6, equation (1.1) has only finitely many solutions in n,d,b,yn,d,b,y."

The paper notes (p. 375) that Darmon and Granville (Bull. London Math. Soc. 27 (1995), 513--543), applying Faltings' theorem, had shown this for b=1b=1, k≥3k\ge3 and l≥4l\ge4 fixed, and that Theorem 6 refines their result and extends it to b>1b>1. It is best possible in the sense that for fixed k≥3k\ge3, l≥2l\ge2 with k+l≤6k+l\le6, (1.1) has infinitely many solutions in each case, citing Tijdeman; and the proof shows the result also holds for solutions of (1.1) with n<0n<0. The theorem gives finiteness only; it gives no bound on the solutions.

Source. K. Győry, L. Hajdu and N. Saradha, On the Diophantine equation n(n+d)⋯(n+(k−1)d)=byln(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 6 and the remarks around it on p. 375, the proof on pp. 385--386. The edition is recorded on the source card.

Read depth. Claims checked: the statement and the remarks were read clause by clause against the published print, and the proof on pp. 385--386 for its structure only. A second reader checked the statement, hypotheses, label and page against the print.

Proof pointer

Section 6, pp. 385--386. Write each term as n+id=aixiln+id=a_ix_i^l as in (3.1) (p. 377), with aia_i free of llth powers and P(ai)≤kP(a_i)\le k, so the aia_i take finitely many values; fix them. Three-term relations such as (n+id)+(n+(i+2)d)=2(n+(i+1)d)(n+id)+(n+(i+2)d)=2(n+(i+1)d) give the identities (6.1)--(6.3), whose product is an equation F(xi,xi+2)=zlF(x_i,x_{i+2})=z^l with FF a binary form having enough pairwise linearly independent linear factors: three such equations multiplied for k≥5k\ge5, two for k=4k=4 (where l≥3l\ge3), one for k=3k=3 (where l≥4l\ge4). Theorem 1 of Darmon and Granville then leaves finitely many values of xix_i and xi+2x_{i+2}, hence of every xix_i, and so of n,d,b,yn,d,b,y.

Dependencies

Theorem 1 of Darmon and Granville (1995), which rests on Faltings' theorem; the factorization (3.1) of the same paper.

Bears on

  • Problem 672: with b=1b=1, for each fixed length k≥4k\ge4 and exponent l≥2l\ge2 with k+l>6k+l>6, that is every l≥2l\ge2 when k≥5k\ge5 and every l≥3l\ge3 when k=4k=4, at most finitely many primitive positive progressions of length kk have a product equal to an llth power. This is finiteness for each fixed pair (k,l)(k,l), not nonexistence; it does not bound the exponent or the length.