# Show that $\psi(n)$ has finitely many roots

Define $$\psi(n)=\pi(n)-\phi(n)$$ where we have the prime counting function and totient function respectively.

I'm interested in where $$\psi(n)=0$$.

Specifically is it possible to prove that there are exactly $$9$$ roots, namely for $$n=0,2,3,4,8,10,14,20,90$$. I calculated these roots but I couldn't find any more.

I think this function has finitely many roots because the totient function grows faster and thus becomes greater than the prime counting function at a certain threshold.

Thanks for your help. I'm not sure how simple or how difficult this is to prove, and I appreciate any ideas and hints.

• Indeed, the ratio explodes and tells you that there are only finitely many solutions. On wikipedia you find upper and lower bounds for those two functions. I would give it a try to see what kind of threshold you get. – Severin Schraven Apr 22 at 0:51

## 1 Answer

yes, Rosser and Schoenfeld showed (formulas 3.41 and 3.42) that $$\phi$$ (which is, on average, linear) is never much worse than that: namely $$\phi(n) \geq \frac{n}{e^\gamma \log \log n + \frac{2.50637}{\log \log n}}$$ Here the logarithm is base $$e$$ and the constant $$2.50637$$ chosen to give equality at (and only at) $$n = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 = 223092870$$

There are also upper bounds on $$\pi(x),$$ for example formula 3.2.