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Source. Theorem 1, p. 12, of Jean-Marie De Koninck, Florian Luca and Igor E. Shparlinski, Powerful numbers in short intervals, Bull. Austral. Math. Soc. 71 (2005), 11--16, doi:10.1017/S0004972700037953. See the source card.
Read depth. Claims checked: the statement was read clause by clause on the printed page. The proof (pp. 12--14) was read for structure only. Nothing here is independently reviewed.
Statement
For an integer , an integer is -full when for every prime dividing (p. 11); the -full integers are the squarefull (powerful) ones.
Theorem 1 (p. 12). For every integer there are infinitely many such that the open interval contains at least
-full integers.
Here is fixed, and the term may depend on it (p. 14). The paper's closing remarks (pp. 15--16) say that running the argument with Liouville's theorem instead of Roth's gives a version explicit and uniform in , with replaced by ; in particular, for infinitely many the interval then holds at least -full integers whenever , and arbitrarily many when . The remarks give no separate proof of these variants.
Proof pointer
Section 2 (pp. 12--14). Take the first squarefree integers above , their product , and . A common denominator , at least an explicit depending on and , approximates all the simultaneously to within ; Dirichlet's simultaneous approximation theorem supplies such a , and Roth's theorem applied to bounds the least one. With the distinct -full numbers all lie within of , so one of and holds at least of them. The size bound then converts into the stated count, being arbitrary.
Dependencies
Roth's theorem and Dirichlet's simultaneous approximation theorem, both cited from W. M. Schmidt, Diophantine approximation (Springer, 1980), Theorem 2A of Chapter 5 and Theorem 1A of Chapter 2.
Bears on
- Problem 942: the case gives, for infinitely many , at least powerful integers in , hence in . This is a lower bound for infinitely many only; it gives no upper bound valid for all and does not settle the problem. The paper does not mention the problem.