Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Statement
Let , the range of Ecklund's theorem, and suppose that has no prime divisor . Then
The printed lemma (p.267) states only the condition on prime divisors; the proof below uses .
Proof
Every prime divisor of is greater than . Such a prime does not divide , because , and it has at most one multiple among . It must therefore occur to exponent one in the numerator interval . In particular , and
Taking logarithms of the product gives , proving the first inequality in (6). Each of the primes in the product is at most , which proves the second.
Verification record
Current review state. Accepted by independent mathematical review, retained as the full-proof review and its final receipt. Substantive changes to this proof or its premises invalidate the affected scope until rechecked.
Scope and source version. The checked scope is equation (6), the standing condition , and the expanded multiplicity-one product argument. The source is Ecklund's Pacific Journal of Mathematics 29 (1969), 267--270 publisher PDF, identified on the source card: the statement is on printed p.267 / physical p.2 and the source proof is on printed p.268 / physical p.3.
Premises and limits. This component uses no external theorem. No gap remains inside the rewritten argument at the accepted scope. No formal verification is recorded.