The following question uses standard sieve theory terminology. Let $A=\vert\{a_n: a_n=n(n-2); n\in[N/2,N]\}\vert$ and let $A_d=\vert\{a_n: d\vert a_n\}\vert$. If we are looking for $S^T$ the number of $a_n$ in $A$ such that $a_n$ has no prime factors less than $\sqrt{N}$, we can use Selberg Sieve to find a nice upper bound for $$S^T=\sum_{d\vert P_z}\mu(d)A_d=\sum_{d\vert P_z}\mu(d)\frac{N/2}{d}\omega(d)+\sum_{d\vert P_z}\mu(d)r_d$$ Where $r_d<\omega(d)$. Now, what I need to show is that $$-a\sum_{d\vert P_z}\mu(d)\frac{N/2}{d}\omega(d)+\sum_{d\vert P_z}\mu(d)R_d <\frac{\epsilon N}{\log^2{N}}$$ For any $\epsilon>0$ and for fixed $a>0$. Please tell me if this seems right: I want to use Selberg Sieve to show this. Using Selberg Sieve, we find a new function to replace the Möbius function, called $\lambda$. The way Selberg set it up, was (I think) by showing that $$\sum_{d\vert P_z}\mu(d)\frac{N/2}{d}\omega(d)\leq \sum_{d\vert P_z}\lambda(d)\frac{N/2}{d}\omega(d)$$
$$\sum_{d\vert P_z}\mu(d)R_d\leq \sum_{d\vert P_z}\lambda(d)r_d$$ Then basically we have that the first term in the first inequality is bigger or equal to zero and less than some multiple of $N/\log^2{N}$, and the second term (using ONLY the fact that $\vert r_d\vert<\omega(d)$) is overwhelmed by that first term thus we can say that $$-a\sum_{d\vert P_z}\mu(d)\frac{N/2}{d}\omega(d)\leq 0$$ and thus $$-a\sum_{d\vert P_z}\mu(d)\frac{N/2}{d}\omega(d)+\sum_{d\vert P_z}\mu(d)R_d\leq\sum_{d\vert P_z}\mu(d)R_d $$ $$\leq 3\sum_{d\vert P_z}\mu(d)r_d < \frac{\epsilon N}{\log^2{N}}$$ For any $\epsilon>0$. It's a little confusing, and only (I think) works because Selberg used a term by term comparison of main term to main term and error term to error term as well as only an upper bound on the error term to bound it. I'm not super certain it is right though.
BOOO I'm an idiot!!! I'm sorry for wasting the time of anyone who may have read this. Though it is true that both error and main term under the influence of the Möbius function and added together are less than when under the influence of the lambda function, that does not mean that the error term in the case of the Möbius function is less than the error term in the case of the lamda function...The above is completely incorrect.