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Source and scope. The estimate used in the proof of Croot's Corollary to Theorem 1, p. 235, is expanded here. An integer is powerful if every prime dividing it has exponent at least two; $1$ is included.

Statement. Write $\mathcal P$ for the powerful positive integers. Then

$$\#\{a\in\mathcal P:a\le t\}=O(\sqrt t).$$

Uniformly for $3/4\le\sigma\le1$ and $Y\ge1$,

$$\sum_{\substack{a\in\mathcal P\\a>Y}}a^{-\sigma} =O(Y^{1/2-\sigma}).$$

Consequently the number of integers $n\le x$ whose powerful part exceeds $Y$ is $O(x/\sqrt Y)$.

Complete proof. Every powerful integer has a unique representation

$$a=b^2d^3,\qquad d\text{ squarefree}.$$

Indeed, an even prime exponent $e$ contributes $p^{e/2}$ to $b$ and nothing to $d$. An odd exponent $e\ge3$ contributes $p^{(e-3)/2}$ to $b$ and $p$ to $d$. Thus

$$\#\{a\in\mathcal P:a\le t\} \le\sum_{d\ge1}\left\lfloor\frac{\sqrt t}{d^{3/2}}\right\rfloor \le\sqrt t\sum_{d\ge1}d^{-3/2}=O(\sqrt t).$$

Partition the tail into $2^jY<a\le2^{j+1}Y$ for $j\ge0$. Its contribution on this interval is at most a constant times

$$(2^{j+1}Y)^{1/2}(2^jY)^{-\sigma}.$$

The resulting geometric series is bounded uniformly because $2^{1/2-\sigma}\le2^{-1/4}<1$. This proves the weighted tail.

Finally, write $n=\alpha\beta$ uniquely by putting each full prime power of exponent at least two into $\alpha$, and each prime of exponent one into $\beta$. Then $\alpha$ is powerful, $\beta$ is squarefree, and $\gcd(\alpha,\beta)=1$. Discarding restrictions on $\beta$ gives

$$\#\{n\le x:\alpha(n)>Y\} \le\sum_{\substack{\alpha\in\mathcal P\\\alpha>Y}} \left\lfloor\frac x\alpha\right\rfloor \le x\sum_{\substack{\alpha\in\mathcal P\\\alpha>Y}}\alpha^{-1} =O(x/\sqrt Y).$$

Source clarification. This argument proves the tail estimate directly. It does not assume that every powerful integer greater than $Y$ has a square divisor greater than $Y$: a prime cube shows why that assumption would fail.

Bears on. the general-modulus corollary and prime-power smoothness.