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Source. The conclusion following equation (2) in Croot's published paper, pp. 233–234. The proof below supplies a compilation repair of the displayed square-divisor comparison, while proving its required asymptotic conclusion.

Use $T(x)=\sqrt{\log x\log\log x}$ and $L(c,x)=e^{cT(x)}$. Define

$$\psi^*(x,y)= \#\{1\le n\le x:p^a\mid n,\ p\text{ prime},\ a\ge1 \Longrightarrow p^a\le y\}.$$

Statement. For every fixed $c>0$ and $y=L(c,x)$,

$$0\le\psi(x,y)-\psi^*(x,y) \le x\exp\left(-\left(\frac1{2c}+\frac c2+o(1)\right)T(x)\right).$$

In particular, the external estimate in Lemma 1 gives

$$\psi^*(x,L(c,x)) =x\exp\left(-\left(\frac1{2c}+o(1)\right)T(x)\right).$$

Complete relative proof. Put $X=\log x$ and

$$u=\frac{\log x}{\log y} =\frac1c\sqrt{\frac X{\log X}},\qquad \delta=\frac{\log u-2\log\log u}{\log y},\qquad \sigma=1-\delta.$$

For sufficiently large $x$ (depending on $c$), $u>1$, $\delta>0$, and $3/4\le\sigma<1$. The elementary Euler-product estimate gives

$$\log Z_y(\sigma) \le y^\delta O(\log\log y)+O(1) =O\left(\frac{u}{\log u}\right)=o(T(x)).$$

Here $y^\delta=u/(\log u)^2$ and $\log\log y/\log u\longrightarrow1$. Also

$$u\log u=\left(\frac1{2c}+o(1)\right)T(x), \qquad u\log\log u=o(T(x)).$$

It follows that

$$x^\sigma Z_y(\sigma) =x\exp\left(-\left(\frac1{2c}+o(1)\right)T(x)\right).$$

For every real $z\ge0$, positivity of the finite Euler-product series gives the elementary Rankin bound

$$\psi(z,y)\le z^\sigma Z_y(\sigma).$$

For $z<1$ the left side is zero; otherwise this follows by bounding each $1$ for a smooth $m\le z$ by $(z/m)^\sigma$.

An integer counted by $\psi(x,y)$ but not by $\psi^*(x,y)$ has a divisor $p^a>y$ with $a\ge2$ and $p\le y$. Dividing by this prime power leaves a $y$-smooth integer. Therefore

$$\begin{aligned} \psi(x,y)-\psi^*(x,y) &\le\sum_{\substack{p\le y,\ a\ge2\\p^a>y}}\psi(x/p^a,y)\\ &\le x^\sigma Z_y(\sigma) \sum_{\substack{p\le y,\ a\ge2\\p^a>y}}p^{-a\sigma}. \end{aligned}$$

Each prime power in the last sum is a distinct powerful integer. The uniform powerful-number tail bounds the sum by

$$O(y^{1/2-\sigma}) =\exp\left(-\left(\frac c2+o(1)\right)T(x)\right),$$

because $\delta\log y=O(\log u)=o(T(x))$. This proves the first display. Compared with Lemma 1, the error has relative size $\exp(-(c/2+o(1))T(x))$, which tends to zero. Subtraction proves the prime-power smoothness estimate.

Why the source comparison is replaced. The sum over square divisors $m^2>y$ in (2) does not directly cover all forbidden prime powers. For example, $8$ has a prime-power divisor exceeding $7$ but no square divisor exceeding $7$. Indeed $\psi(8,7)=8$, $\psi^*(8,7)=7$, and the displayed square-divisor sum is zero. This finite example identifies the invalid general comparison; it is not a counterexample to the asymptotic theorem. The union bound over actual prime powers above closes the required argument.

Dependencies. Lemma 1 is external. The Euler-product and powerful-tail estimates are proved in the linked pages; all subsequent deductions are included here.

Bears on. the lower-bound construction and Problem 202.