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Statement

Lemma 4.3 (printed p. 5). "n!+kn!+k is powerful finitely often."

The lemma is one of the parts into which the paper breaks the proof of Theorem 4.1, so kk is the fixed integer k≥0k\ge0 of that theorem, and the statement is read as: for fixed kk, only finitely many nn make n!+kn!+k powerful. The lemma's statement does not mention the abc conjecture; its proof invokes it, so the lemma holds as proved only under the hypothesis of Theorem 4.1. The proof takes n≥kn\ge k and uses k∣n!k\mid n!, so it treats k≥1k\ge1; the case k=0k=0 is Lemma 4.2.

Source. D. Cushing and J. E. Pascoe, Powerful numbers and the ABC-conjecture, arXiv:1611.01192v1 (3 November 2016); Lemma 4.3 on p. 5, its proof on pp. 5--6. The edition is identified in the source digest.

Read depth. Claims checked: the statement and the shape of the proof were read on the page images of the preprint; the inequalities were not checked step by step, and nothing here is independently reviewed.

Proof pointer

Pp. 5--6. For n≥kn\ge k and n!+k=xn!+k=x with xx powerful, the proof divides through by kk and applies the abc conjecture with ε=12\varepsilon=\frac12 to the coprime triple n!/k+1=x/kn!/k+1=x/k. The radical of the product is at most n#⋅x1/2n\#\cdot x^{1/2}, by rad⁡(n!)=n#\operatorname{rad}(n!)=n\# (Lemma 2.5, p. 3) and the radical bound for powerful numbers (Lemma 2.6, p. 3, used in the form rad⁡(x)≤x1/2\operatorname{rad}(x)\le x^{1/2}). Because n#n\# grows only exponentially in nn while x>n!x>n!, the triple satisfies rad⁡(abc)3/2<c\operatorname{rad}(abc)^{3/2}<c for all large nn, which the abc conjecture allows only finitely often. The final display writes x=cx=c where the triple has c=x/kc=x/k.

Bears on

  • Problem 936: with k=1k=1 the lemma gives, assuming abc, that n!+1n!+1 is powerful for only finitely many nn. This is the n!+1n!+1 case of the problem, conditionally; the n!−1n!-1 case is Exercise 4.4, left to the reader.
  • Problem 398: every square is powerful, so the case k=1k=1 also gives, assuming abc, only finitely many solutions of n!+1=x2n!+1=x^2. The problem asks whether the known solutions are the only ones, which a finiteness statement does not answer; the paper's introduction (p. 2) recalls this finiteness as already shown under abc by Overholt (its reference [2]).