Expressing $\pi(x) = \frac{P(x)}{Q(x)}$, where $P$ and $Q$ are polynomials

Question

In Tom Apostol's Analytic Number Theory book there is a problem which states:

• That there do not exists polynomials $P(x)$ and $Q(x)$ such that $$\pi(x) = \frac{P(x)}{Q(x)}$$ for all $x \in \mathbb{N}$.

I tried this problem, but couldn't find a solution. Here is what i attempted. Since $\pi(x) \sim x\log{x}$ as $x \to \infty$, i saw what happens to the Right hand side $\frac{P(x)}{Q(x)}$ as $x \to \infty$. I couldn't conclude anything. If there are interesting proof's for this result, i shall be happy to see it.

Answer 1

It seems like you've already gotten the answer. By asymptotic considerations, we must have $\deg P(x) > \deg Q(x)$. Moreover, if we have $\deg P(x) = \deg Q(x) + 1$, then we have $$ \frac{\pi(x) Q(x)}{P(x)} \sim \log x \frac{xQ(x)}{P(x)}, $$ but the $xQ(x)/P(x)$ is asymptotic to the ratio of the leading coefficients, so this can't be asymptotic to $1$, i.e. $P(x)/Q(x)$ grows too slowly. On the other hand, if $\deg P(x) = \deg Q(x)+2$, then we similarly see that $P(x)/Q(x)$ grows too quickly to be asymptotic to $x \log x$.

Answer 2

HINT $\ $ A rational function is asymptotic to $\rm\ x^n,\ $ for some $\rm\ n \in \mathbb Z\:.\ $ But $\rm\ x \log(x)\ $ lies strictly between $\rm\ x\ $ and $\rm x^2$ asymptotically, i.e. $\rm\ (x\ log(x))/x = log(x)\to\infty,\ \ x^2/(x log(x)) = x/log(x)\to \infty$