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Source. Croot, published paper, p. 234, the construction preceding Theorem 1. This refines the Erdős–Szemerédi construction cited there by allowing prime-power factors.

Let $f(x)$ be the maximum number of pairwise disjoint integer congruence classes with distinct moduli in $[2,x]$. Write $T(x)=\sqrt{\log x\log\log x}$.

Statement. For each $\eta>0$ and all sufficiently large $x$,

$$f(x)\ge x\exp(-(\sqrt2+\eta)T(x)).$$

Complete relative proof. Put $c=1/\sqrt2$, and let $p$ be the largest prime at most $e^{cT(x)}$. Bertrand's postulate gives

$$\log p=cT(x)+O(1).$$

Take every integer $m\le x/p$ all of whose prime-power divisors are strictly less than $p$, and use the modulus $q=pm$. In particular $p\nmid m$. Write the maximal prime-power factors of $m$ in decreasing order as

$$d_r>d_{r-1}>\cdots>d_1, \qquad m=d_r\cdots d_1.$$

The $d_j$ are powers of distinct primes, and all satisfy $d_j<p$. Define a residue $a_m\pmod{pm}$ by

$$a_m\equiv d_r\pmod p,\qquad a_m\equiv d_{j-1}\pmod{d_j}\ (2\le j\le r),\qquad a_m\equiv0\pmod{d_1}.$$

The Chinese remainder theorem applies because the listed moduli are pairwise coprime. When $m=1$, define $a_1\equiv0\pmod p$ instead.

To prove disjointness, associate the descending list $(p,d_r,\ldots,d_1)$ to $m$. For two distinct integers $m,m'$, their lists have a longest common initial segment, which contains $p$. Let $D$ be its last entry. The next entries of the two lists are different; if a list has ended, use $0$ for its next entry. Both entries lie in $[0,D)$, and the construction makes them the respective residues of $a_m$ and $a_{m'}$ modulo $D$. Thus these residues differ modulo $D$, although $D$ divides both moduli. An integer cannot belong to both congruence classes.

It remains to count the moduli. Their number is

$$\psi^*(x/p,p-1).$$

Set $x'=x/p$. Since $\log p=O(T(x))=o(\log x)$,

$$\frac{T(x')}{T(x)}\longrightarrow1,\qquad \frac{\log(p-1)}{T(x')}\longrightarrow c.$$

For every fixed $0<\varepsilon<c$, the second relation eventually places $p-1$ between $L(c-\varepsilon,x')$ and $L(c+\varepsilon,x')$. Monotonicity in the smoothness cutoff and the complete prime-power smoothness deduction therefore imply, by letting $\varepsilon$ tend to zero after taking limits,

$$\begin{aligned} \psi^*(x/p,p-1) &=\frac xp\exp\left(-\left(\frac1{2c}+o(1)\right)T(x)\right)\\ &=x\exp\left(-\left(c+\frac1{2c}+o(1)\right)T(x)\right)\\ &=x\exp(-(\sqrt2+o(1))T(x)). \end{aligned}$$

The constructed distinct moduli give the desired lower bound for $f(x)$.

Source clarification. The phrase on p. 234 requiring both divisibility by $p$ and all prime-power factors to be less than $p$ must refer to the factors of $q/p$. Taken literally for $q$, it would exclude every modulus. The formula $q=p\,d_r\cdots d_1$ and the subsequent CRT conditions identify the intended meaning. The empty list and the disjointness argument are written out above.

Dependencies. The smooth-number theorem in Lemma 1 remains an external analytic input. Bertrand's postulate and the Chinese remainder theorem are classical inputs; the construction and all counting deductions specific to this paper are included.

Bears on. Problem 202: a lower bound for its maximum with coefficient $\sqrt2$ on the scale $T(x)$, not the sharp coefficient.