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Source. Part II, display (3) and its following proof, printed page 200 (PDF page 4), of Erdős (1974).
Let be as in threshold_comparison, and put
Statement. There is an absolute such that
for infinitely many integers . More generally, every satisfies
External input. Prachar's Satz 2 gives a constant and infinitely many for which the number of odd primes with exceeds
The proof of that analytic theorem is not included here.
Complete deduction. Choose an admissible pair . If divided neither nor , Fermat's theorem would imply , contradicting coprimality. Thus every such prime divides , and the primes are distinct, so
To obtain the asymptotic conclusion, take the infinite sequence supplied by Prachar. Along this sequence . The product of any distinct primes is at least , and for all sufficiently large . For completeness, at least factors in are at least , so eventually. Hence
Exponentiating gives the claim with . 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 of primes, print an extra exponential: they have in place of . The latter is the input actually stated in Prachar's original Satz 2, printed page 91. The former eventually exceeds , whereas is at most the number of divisors of , hence at most . 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- lower bound and does not answer whether infinitely often.