Wiki
Wiki

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

Updated


Source. Part II, display (3) and its following proof, printed page 200 (PDF page 4), of Erdős (1974).

Let H(n)H(n) be as in threshold_comparison, and put

P(n)={p:p prime, p−1∣n},s(n)=∣P(n)∣.\mathcal P(n)=\{p:p\text{ prime},\ p-1\mid n\},\qquad s(n)=|\mathcal P(n)|.

Statement. There is an absolute c>0c>0 such that

H(n)>exp⁡ ⁣(nc/(log⁡log⁡n)2)H(n)>\exp\!\left(n^{c/(\log\log n)^2}\right)

for infinitely many integers nn. More generally, every n≥2n\ge2 satisfies

H(n)2>∏p∈P(n)p.H(n)^2>\prod_{p\in\mathcal P(n)}p.

External input. Prachar's Satz 2 gives a constant c0>0c_0>0 and infinitely many nn for which the number of odd primes pp with p−1∣np-1\mid n exceeds

exp⁡ ⁣(c0log⁡n(log⁡log⁡n)2)=nc0/(log⁡log⁡n)2.\exp\!\left(c_0\frac{\log n}{(\log\log n)^2}\right) =n^{c_0/(\log\log n)^2}.

The proof of that analytic theorem is not included here.

Complete deduction. Choose an admissible pair 2≤a<b=H(n)2\le a<b=H(n). If p∈P(n)p\in\mathcal P(n) divided neither aa nor bb, Fermat's theorem would imply an≡bn≡1(modp)a^n\equiv b^n\equiv1\pmod p, contradicting coprimality. Thus every such prime divides abab, and the primes are distinct, so

H(n)2=b2>ab≥∏p∈P(n)p.H(n)^2=b^2>ab\ge\prod_{p\in\mathcal P(n)}p.

To obtain the asymptotic conclusion, take the infinite sequence supplied by Prachar. Along this sequence s=s(n)→∞s=s(n)\to\infty. The product of any ss distinct primes is at least s!s!, and s!>e2ss!>e^{2s} for all sufficiently large ss. For completeness, at least ⌊s/2⌋\lfloor s/2\rfloor factors in s!s! are at least s/2s/2, so log⁡(s!)≥⌊s/2⌋log⁡(s/2)>2s\log(s!)\ge\lfloor s/2\rfloor\log(s/2)>2s eventually. Hence

log⁡H(n)>12log⁡ ⁣(∏p∈P(n)p)>s(n)>nc0/(log⁡log⁡n)2.\log H(n)>\frac12\log\!\left(\prod_{p\in\mathcal P(n)}p\right)>s(n) >n^{c_0/(\log\log n)^2}.

Exponentiating gives the claim with c=c0c=c_0. This elementary factorial estimate replaces the source's prime-number-theorem estimate; the core Fermat/product argument is the same.

Source correction. The prose immediately after (3), and the subsequent bound on its number ss of primes, print an extra exponential: they have exp⁡(nc2/(log⁡log⁡n)2)\exp(n^{c_2/(\log\log n)^2}) in place of nc2/(log⁡log⁡n)2n^{c_2/(\log\log n)^2}. The latter is the input actually stated in Prachar's original Satz 2, printed page 91. The former eventually exceeds nn, whereas s(n)s(n) is at most the number of divisors of nn, hence at most nn. Reading the correct input proves the unchanged conclusion (3). The reference year is also 1955 in the original volume, not 1954 as listed on Erdős's page 200. Prachar's submission date was 19 October 1954.

Dependencies. The exact external Prachar theorem, Fermat's little theorem, elementary divisibility, and the definitions in threshold_comparison. This is a complete relative proof of (3), not a proof of Prachar's theorem or of the conjectured optimal constant.

Bears on. #820. Later stronger shifted-prime-divisor estimates can be substituted into the same product argument. The infinitely-often conclusion provides no all-large-nn lower bound and does not answer whether H(n)=3H(n)=3 infinitely often.