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Let $a$ be a positive number. Calculate the sum $$\sum_{1\le n\le x}\left\lfloor \sqrt{n^{2}+a} \right\rfloor$$

I tried to calculate first $\left\lfloor \sqrt{n^{2}+a} \right\rfloor-n$. But probably it won't help. Maybe you have some ideas how to solve it?

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