Let $(\frac{m}{n})$ be the Jacobi quadratic symbol defined for positive squarefree odd integers $n,m$. Does the following sum go to infinity? $$ \sum_{1\leq n \leq (\log x)^{100} } \mu^2(2n) \sum_{(\log x )^{100} < m \leq x} \left(\frac{m}{n}\right) \frac{\mu^2(2 m )}{m}, $$ where $\mu$ is the M"obius function. The sum over $m$ looks as the tail of $L(1,(\frac{\cdot}{n}))$ where $L$ is the Dirichlet $L$-function associated to the quadratic character modulo $n$.
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If in doubt use the square-root cancellation heuristic. You can think of this sum as $$ \sum_{n < (\log x)^{100}} \varepsilon_n X_n $$ where $\varepsilon_n$ is a random sign and $X_n$ is another random variable usually of size $$ \sqrt{\text{Var}(X_n)} := \sqrt{\sum_{m > (\log x)^{100}} \frac{1}{m^2}} \asymp (\log x)^{-100/2}. $$ Then you expect that the optimal bound for the entire sum to be $$ \sqrt{\sum_{n < (\log x)^{100}} \text{Var}(X_n)} \asymp 1 $$ However based on probabilistic heuristics (e.g law of iterated logarithm) you'd expect there to be some additional fluctuations of the order of $\sqrt{\log\log\log x}$ so I'd guess that this sum can go to infinity on a subsequence.
