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Source. The theorem on printed p. 143, proved on p. 147 (PDF pp. 1 and 5).

Let $f(x)$ be the maximum size of a pairwise disjoint family of residue classes with distinct moduli in $[2,x]$. For every fixed $\eta>0$,

$$f(x)\le x\exp\left(-\left(\frac12-\eta\right) \sqrt{\log x\log\log x}\right)$$

for all sufficiently large $x$. The assertion is unconditional and allows arbitrary prime powers in the moduli. It is a historical bound, not the current sharp answer to Problem 202.

Complete proof

It suffices to prove the assertion for $0<\eta<1/2$. Put $T=\sqrt{\log x\log\log x}$. Choose a fixed integer $k\ge4$ with $4/k<\eta/3$, and then a fixed $\delta>0$ with $2\delta<\eta/3$. Set $r=k(k+1)$. These choices precede the limit $x\to\infty$.

Take any admissible family and let $\mathcal U$ consist of its moduli $n$ with $h_r(n)<e^{2T}$. By Lemma 4, the complementary family has size at most

$$3x\exp\left(-2\left(1-\frac2k\right)T\right)\le3xe^{-T}.$$

The counting step in that same lemma shows that there are at most

$$(e^{2T})^{2/k}=e^{4T/k}$$

possible values of $h_r(n)$ in $\mathcal U$.

If $\mathcal U$ is nonempty, some fixed value $a<e^{2T}$ therefore occurs in at least $|\mathcal U|e^{-4T/k}$ moduli. A further pigeonhole retains at least

$$\frac{|\mathcal U|}{a e^{4T/k}}$$

original residue classes having the same residue modulo $a$. Write their moduli as $n_i=a d_i$. Their $d_i=l_r(n_i)$ are distinct, coprime to $a$, at most $x/a$, and have every prime exponent at most $r$.

Dropping the factor $a$ preserves disjointness of the classes $b_i\pmod{d_i}$. Indeed, if two of these had a common integer, their congruences would specify a class modulo $\operatorname{lcm}(d_i,d_j)$. That modulus is coprime to $a$. The Chinese remainder theorem could combine this class with the shared original residue modulo $a$, producing an intersection of the two original classes, a contradiction.

There is one minor endpoint: if a retained $d_i=1$, its reduced class is all of $\mathbb Z$, so the reduced disjoint family has only one member. Thus the bound in Lemma 6 still applies for all large $z=x/a$: the theorem’s right side tends to infinity and absorbs a singleton. Otherwise all $d_i\ge2$ and the lemma applies directly.

Uniformly for $1\le a<e^{2T}$,

$$z=x/a\ge xe^{-2T}\longrightarrow\infty,\qquad \frac{\sqrt{\log z\log\log z}}T\longrightarrow1.$$

To see the second limit, $\log z/\log x=1+O(T/\log x)\to1$, and consequently $\log\log z/\log\log x\to1$, uniformly. Lemma 6 with the fixed $r,\delta$ therefore gives, for all large $x$,

$$\frac{|\mathcal U|}{a e^{4T/k}} \le \frac xa \exp\left(-\left(\frac12-\delta\right) \sqrt{\log(x/a)\log\log(x/a)}\right) \le \frac xa e^{-(1/2-2\delta)T}.$$

The choice of one uniform large-$x$ threshold is legitimate: $r,\delta$ are fixed and the smallest possible $z$ tends to infinity. Hence

$$|\mathcal U|\le x e^{-(1/2-4/k-2\delta)T}.$$

This also holds when $\mathcal U$ is empty. Adding the complementary $3xe^{-T}$ term and using $4/k+2\delta<2\eta/3$ proves the result.

The explicit coprimality argument, singleton case and uniform $x/a$ rescaling expand the abbreviated source deduction. The source’s six-page method is fully reconstructed at its stated external smooth-number input and canonical Croot proof boundaries. It neither evaluates the later sharp constant nor settles an additional covering problem.