GPT 6 Astra Ultra attempt on Math-5525
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Contents
Erdős Problem 240: an infinite prime set with separated smooth numbers
1) Statement and scope
Is there an infinite set of primes $P$ such that successive integers composed only of primes in $P$ have gaps tending to infinity? We reconstruct Tijdeman's inductive argument, obtaining the stronger bound $b-a>\sqrt a$ for every two such integers $a<b$. A consequence of Baker's theorem on logarithms is the explicit deep input.2) The external logarithmic estimate
The consequence needed here is as follows. Fix a finite prime set $P_0$. There exists $C=C(P_0)$ such that, for any new prime $q>\max P_0$, if \[ a=q^\alpha\prod_{p\in P_0}p^{u_p},\qquad b=q^\beta\prod_{p\in P_0}p^{v_p},\qquad \alpha\ne\beta,\qquad 0<b-a\leq\sqrt a, \] then $a,b\leq q^C$. In particular $\alpha,\beta\leq C$ and $u_p,v_p\leq C\log q/\log p$. This is the estimate derived from Baker's theorem in the proof of Tijdeman's Theorem 7, with its parameter $\theta=1/2$ and with the constant enlarged if necessary. Its crucial uniformity is that $C$ is independent of $q$. We use this estimate as an external theorem, not as an unproved elementary assertion.3) Extending a finite prime set
Assume all $P_0$-smooth pairs $a<b$ satisfy $b-a>\sqrt a$. Let $q\in[T/2,T]$ be a candidate new prime. If a bad pair in $P_0\cup\{q\}$ has $\alpha=\beta$, division by $q^\alpha$ gives a pair $a',b'$ from the old semigroup, and \[ b-a=q^\alpha(b'-a')> q^\alpha\sqrt{a'}\geq\sqrt a, \] a contradiction. Thus every bad pair has $\alpha\ne\beta$. By the external estimate, the number of possible exponent tuples is $O_{P_0}((\log T)^{2|P_0|})$.Fix one such tuple and write $b/a=cq^d$, with $d\ne0$ fixed. At least one of $a,b$ is divisible by $q$, so $b\geq q\geq T/2$. Since $b\leq a+\sqrt a\leq2a$, we have $a\geq T/4$. Consequently \[ 0<\log c+d\log q=\log(b/a)\leq2T^{-1/2}. \] The allowed values of $\log q$ lie in an interval of length at most $2T^{-1/2}/|d|$. Restricted to $q\in[T/2,T]$, this gives an interval of $q$ values of length $O(\sqrt T)$. There are therefore $O(\sqrt T)$ integers, hence at most that many primes, excluded by this tuple.
The total number of bad primes is \[ O_{P_0}\left(\sqrt T(\log T)^{2|P_0|}\right) =o(T/\log T). \] The standard prime-counting estimate $\#\{q\text{ prime}:T/2\leq q\leq T\}\gg T/\log T$ shows that some prime remains when $T$ is sufficiently large. Taking also $T/2>\max P_0$ extends the prime set and preserves the required separation.