# Is $\small \sum\limits_{p\leq x}\pi\left(\frac{x^2}{p}\right)-\sum\limits_{p\gt x}\pi\left(\frac{x^2}{p}\right)=\pi^2(x)$ of any use?

I was playing with some Meissel/Lehmer formulas and I found this one. In fact there is a much simpler way to find it when looking closer, so I guess i is well known.

$$\sum\limits_{p\leq x}\pi\left(\frac{x^2}{p}\right)-\sum\limits_{p\gt x}\pi\left(\frac{x^2}{p}\right)=\pi^2(x)$$

There is a similar formula I already explored in an other context (Goldbach).

$$\sum\limits_{p\leq x}\pi(\small 2x-p \normalsize)-\sum\limits_{p\gt x}\pi(\small 2x-p \normalsize)=\pi^2(x)$$

Any paper covering one of them? Were they used anywhere?

Edit: this seems to work also for $$\sum\limits_{p\leq x}\pi\left(\small\sqrt[k]{2 x^k-p^k}\normalsize\right)-\sum\limits_{p\gt x}\pi\left(\small\sqrt[k]{2 x^k-p^k}\normalsize\right)=\pi^2(x)$$ with $$k\ge 1$$

and for some $$k\gt \gamma_x$$ we even have

$$\sum\limits_{p}\pi\left(\small\sqrt[k]{2 x^k-p^k}\normalsize\right)=\pi^2(x)$$

To illustrate the influence of $$k$$, here is a chart with $$k=1$$ in blue, $$k=2$$ in purple and $$k=16$$ in yellow. $$x=9$$ in this example: wrapping arround $$\pi^2(x)$$

I guess there are a lot of other working formulas...

• No because $\pi(x)^2 = \sum_{p \le x, q \le x} 1 = \sum_{pq \le x^2} 1 - \sum_{pq \le x^2, p > x} 1-\sum_{pq \le x^2, q > x} 1=\sum_{pq \le x^2} 1 -2\sum_{pq \le x^2, p > x} 1$ $=\sum_{pq \le x^2,p \le x} 1 -\sum_{pq \le x^2, p > x} 1 =\sum_{ p \le x}\pi(x^2/p)-\sum_{ p>x} \pi(x^2/p)$ is valid for any set of strictly positive integers $A$ and $\pi(x) = \sum_{n \le x, n \in A} 1$ – reuns Jan 14 at 16:22
• That was my guess. It was not very usefull on Goldbach either. – Collag3n Jan 14 at 18:04