Number of square-full numbers less than $x$ is $\ll \sqrt{x}$ A square-full number is a positive integer $ m $ such that if a prime $ p $ divides $ m $, then $ p^{2}|m $. I would like to prove the bound $$ |\{n \leq x| n \text{ square-full } \}| \ll x^{\frac{1}{2}} $$ I know that the square-full numbers are exactly does of the form $ a^{2}b^{3} $ but I haven't succeeded in proving the stated bound using this. 
I appreciate any suggestions and comments! Thank you!
 A: You are correct that the square-full numbers are of the form $a^2b^3$.  The number of such numbers less than $x$ is less than $\sqrt x + \sqrt {\frac x8} + \sqrt {\frac x{27}} + \dots +\sqrt {\frac x{n^3}}$ where we should have a floor function on each term.  We have overcounted a bit because terms of the form a^9 are counted in the $(a^3)^2a^3$ form and again in the $(a^3)^3$ form, for example.  Each term comes from a value of $b$ and we stop when $n^3 \gt x$.  This means the number less than $x$ is less than $\sqrt x \sum_{i=1}^{\infty}i^{-3/2}=\sqrt x \zeta(\frac 32) \approx 2.612\sqrt x $
A: You are trying to count all pairs $(a,b)$ of naturals such that $a^2b^3 \le x$. You need to show their number is bounded by $C\sqrt{x}$ for some fixed constant $C$, this is the meaning of $\ll$ according to Vinogradov. 
For a fixed $a$ the number of $b$ is at most $\sqrt[3]{x/a^2}$. 
Thus you need to estimate $\sum_{a=1}^{\lfloor \sqrt{x} \rfloor}\sqrt[3]{x/a^2} \le  \sqrt[3]{x} \sum_{a=1}^{\lfloor \sqrt{x} \rfloor} a^{-2/3} $.
That sum is bounded by  $\int_{0}^{\sqrt{x}} a^{-2/3}da \le 3(\sqrt{x})^{1/3}= 3 x^{1/6}  $. Thus accounting for the additional factor of $\sqrt[3]{x}$ we get $3 x^{1/3 +1/6} = 3 x^{1/2}$ as a bound, and the claim follows.  
