From sum of square of cube function $r_3(n)$, I believed that, as $r_k(n)$ denotes the number of ways that $n$ is the sum of $k$ squares, $$\sum_{n\le X} r_3(n)^2\asymp X^{3/2}\log^{C}X,...(*)$$ but with this image (from https://link.springer.com/article/10.1007/BF01303063?noAccess=true) shouldn't it means $$\sum_{n\le X}r_3(n)^2\asymp X^2?$$
This seems pretty possible to me since $\displaystyle r_4(n)=8\sum_{\substack{d|n \\ 4\not |d}} d\gg\sigma(n),$ thus $$\sum_{n\le X}r_4(n)^2\gg \sum_{n\le X}n^2\asymp X^3.$$
So, is it really possible to attain $(*)$? Or actually we only have the magnitude of $X^2$? I can't grasp the paper but I need to use $(*)$ result and as Prof confirmed it before so I am not sure if I interpret the paper correctly.