Chebyshef's theorem gives that $a f(x) \leq \pi(x) \leq b f(x)$ with $f(x)= \frac{c}{\log (x)}$ and some constants $a,b$. Writting $a \sim b$ if $\lim_{x \to \infty} \frac{a(x)}{b(x)}$ we have $P(x) \sim ax^p$ and $Q(x) \sim bx^q$ for some $a,b,p,q$. If the above equality would hold, $\pi(x) \sim c x^r$ for $c=\frac ab$, $r=p-q$. But it's clear that $\lim_{x \to \infty} \frac{\pi(x)}x =0$ by Chebyshef's theorem, thus we need $r\leq 0$, which again would say that $\pi(x)$ is bounded, surely wrong. It's not difficult to show that any real $r$ is impossible using the other half of Chebyshef's theorem.

We don't really need Chebyshev's bounds. By contradiction, admit there're polynomials $ P, Q \in \mathbb{C}[x]$ such that $ \pi(x) = \frac{P(x)}{Q(x)}$, for any $ x \in \mathbb{N}= \{1, 2, \ldots\}$. Then let $ m = \deg(P)$ and $ n = \deg(Q)$. As $ \lim_{x \to+\infty}\pi(x) =+\infty$, obviously $ m \ge n+1$. Now consider the function $ f: \mathbb{R}^{+}\mapsto \mathbb{C}: x \to P(x)-\pi(x) Q(x)$ and observe that $ f(x) = 0$, whenever $ x \in \mathbb{N}$. Then recall there exists $ k \in \mathbb{N}$ such that each integer in the $ (m+1)$-set $ \{k, k+1, \ldots, k+m\}$ is composite. This means the function $ g: \mathbb{C}\mapsto \mathbb{C}: x \to P(x)-\pi(k) \cdot Q(x)$ has at least $ m+1$ distinct zeros. But actually $ g$ is a complex polynomial of degree $ m \ge 1$ and hence it has at most $ m$ distinct zeros in $ \mathbb{C}$, by Gauss' fundamental theorem of algebra (absurd). So the claim.

Peter wrote: Prove that there do not exist polynomials $ P$ and $ Q$ such that \[ \pi(x)=\frac{P(x)}{Q(x)}\] for all $ x\in\mathbb{N}$. We first prove the following claim. Theorem. Let $ l_{0}\in\mathbb{N}$. Suppose that $ \zeta :\mathbb{N}\rightarrow\mathbb{R}$ be a nonconstant function having the property that, for all $ l\geq l_{0}$, there exists a sequence of distinct positive integers $ a_{1}$, $ \cdots$, $ a_{l}$ with \[ \zeta\left( a_{1}\right)=\cdots=\zeta\left( a_{l}\right).\] Then, there is no rational function $ f$ satisfying $ \zeta(n) = f(n)$ for all $ n\in\mathbb{N}$. Proof. Assume to the contrary that there exist a rational function $ f$ such that $ \zeta(n) = f(n)$ for all $ n\in\mathbb{N}$. Since $ f$ is rational, one may write $ f(x) =\frac{P(x)}{Q(x)}$ for some polynomials $ P(x)$, $ Q(x)\in\mathbb{R}[x]$. Take a sufficiently large positive integer $ l$ so that $ l > max(deg\; P,\; deg\; Q)$. By the assumption, one may write \[ \zeta\left( a_{1}\right)=\cdots=\zeta\left( a_{l}\right)=\lambda\] for some positive integers $ a_{1}$, $ \cdots$, $ a_{l}$ and $ \lambda\in\mathbb{R}$. Introduce the polynomial $ \Omega(x)\in\mathbb{R}[x]$ defined by \[ \Omega(x)=P(x)-\lambda Q(x).\] Since $ \lambda =\zeta( a_{k}) = f\left( a_{k}\right) =\frac{ P( a_{k}) }{ Q( a_{k}) }$ , we find that $ \Omega\left( a_{k}\right) = 0$ for all $ k\leq l$, which implies that the polynomial $ \Omega(x)$ has $ l$ distinct roots. Since $ deg\;\left( P(x)-\lambda Q(x)\right)\leq max(deg\; P,\; deg\; Q) < l$ or $ deg\;\Omega(x) < l$, the fundamental theorem of algebra implies that $ \Omega$ is identically zero so that $ P(x)\equiv\lambda Q(x)$. Since $ f(x) =\frac{P(x)}{Q(x)}$, this implies that $ f(x)\equiv\lambda$. It follows that $ \zeta(n) = f(n) =\lambda$ for all $ n\in\mathbb{N}$ so that $ \zeta :\mathbb{N}\rightarrow\mathbb{R}$ is a constant function. This is a contradiction. Now, we prove the main result. We need to show that the function $ \pi$ satisfies the conditions in the theorem. Proposition. Let $ \pi :\mathbb{N}\rightarrow\mathbb{Z}_{\geq 0}$ be the function defined by \[ \pi(n)=\left\vert\;\{ p\leq n\;\vert\; p\; is\; prime.\}\;\right\vert\] for all $ n\in\mathbb{N}$. Then, $ \pi$ is nonconstant. However, for any $ l\in\mathbb{N}$, there exists a sequence of consecutive positive integers $ c+1$, $ \cdots$, $ c+l$ such that \[ \pi\left( c+1\right)=\cdots=\pi\left( c+l\right).\] Proof. Since there are infinitely many primes, it's immediate that $ \pi$ is not a constant function. The second assertion also follows from the fact that there exist are arbitrary length sequences of consecutive composite numbers. Lemma. Let $ l\in\mathbb{N}$. There are infinitely many $ c\in\mathbb{N}$ such that $ c+1$, $ \cdots$, $ c+l$ are all composite numbers. First Proof. Since there are infinitely many primes, we can pick distinct primes $ p_{1}$, $ \cdots$, $ p_{l}$. Consider the system of linear congruences \[ \begin{cases}x\equiv-1\;\left( mod\; p_{1}\right),\\ x\equiv-2\;\left( mod\; p_{2}\right),\\ \;\;\;\;\;\;\;\;\;\;\;\cdots\\ x\equiv-l\;\left( mod\; p_{l}\right).\\ \end{cases}\] Since $ gcd\left( p_{i}, p_{j}\right) = 1$ for all $ i\neq j$, Chinese Remainder Theorem implies that it has a unique solution modulo $ p_{1}\cdots p_{l}$. Hence, one may take a sufficiently large integer $ c > p_{1}\cdots p_{l}$ so that \[ p_{k}\;\vert\; c+k\] for all $ k\leq l$. Since $ c+k > p_{k}$ and and $ c+k$ is divisible by $ p_{k}$, we find that $ c+k$ is always composite. Second Proof. The $ l$ consecutive integers \[ (l+1)!+2,\cdots, (l+1)!+l+1\] are all composite because $ (l+1)!+k$ is divisible by $ k$ for all $ k\in\{2,\cdots, l+1\}$.

Peter wrote: Prove that there do not exist polynomials $ P$ and $ Q$ such that \[ \pi(x)=\frac{P(x)}{Q(x)}\] for all $ x\in\mathbb{N}$. We prove the following claim. Proposition 1. Suppose that $ \eta :\mathbb{R}\rightarrow\mathbb{R}$ is a function having the property that \[ \lim_{x\rightarrow\infty}\eta(x)=\infty\;\;\text{and}\;\;\lim_{x\rightarrow\infty}\frac{\eta(x)}{x}=0.\] Then, $ \eta$ is not a rational function. Proof. Assume to the contrary that $ \eta(x) =\frac{P(x)}{Q(x)}$, where $ P(x)$, $ Q(x)\in\mathbb{R}[x]$. On the one hand, \[ \lim_{x\rightarrow\infty}\frac{P(x)}{Q(x)}=\infty\] implies that $ \deg P\geq 1+\deg Q$. On the other hand, \[ \lim_{x\rightarrow\infty}\frac{P(x)}{xQ(x)}=0\] implies that $ \deg P\leq\deg Q$. This is a contradiction. Now, we establish that the function $ \pi$ satisfies the conditions. Proposition 2. Let $ \pi :\mathbb{R}_{\geq 0}\rightarrow\mathbb{Z}_{\geq 0}$ be the function defined by \[ \pi(x)=\left\vert\;\{ p\leq x\;\vert\; p\; is\; prime.\}\;\right\vert\] for all $ x\in\mathbb{R}_{\geq 0}$. Then \[ \lim_{x\rightarrow\infty}\pi(x)=\infty\;\;\text{and}\;\;\lim_{x\rightarrow\infty}\frac{\pi(x)}{x}=0.\] Reference. A Classical Introduction to Modern Number Theory by Kenneth Ireland, Michael Rosen Lemma 1. When $ x\geq 1$, we have \[ \theta(x)\equiv\sum_{\substack{ p: prime,\\ p\leq x}}\ln p\leq\left( 4\ln 2\right) x.\] Proof Since $ \binom{2n}{n}=\frac{(n+1)\cdots (2n) }{n!}$ is divisible by all primes $ p$ with $ p\in (n, 2n)$, we get \[ \prod_{\substack{ p: prime,\\ n<p<2n}}p\leq\binom{2n}{n}\leq\sum_{k=0}^{2n}\binom{2n}{k}\leq 2^{2n}.\] or \[ \theta(2n)-\theta(n)\leq 2n\ln 2.\] Since $ x\geq 1$, we can find a positive integer $ k$ so that $ x\in\left(2^{k-1}, 2^{k}\right]$. It follows that \[ \theta(x)\leq\theta( 2^{k})=\sum_{i=1}^{k}\left(\theta( 2^{i})-\theta( 2^{i-1})\right)\leq\sum_{i=1}^{k}2^{i}\ln 2=\left( 2^{k+1}-2\right)\ln 2 < 2^{k+1}\ln 2 <\left(4\ln 2\right)x.\] Lemma 2. When $ x\geq 2$, we have \[ \pi(x)\leq\left( 2+8\ln 2\right)\frac{ x }{\ln x}.\] Proof Let $ x\geq 2$. It follows that \[ \theta(x)\geq\sum_{\substack{ p: prime,\\ \sqrt{x}<p\leq x}}\ln p\geq\sum_{\substack{ p: prime,\\ \sqrt{x}<p\leq x}}\ln\sqrt{x}=\left(\pi(x)-\pi(\sqrt{x})\right)\ln\sqrt{x}\geq\pi(x)\ln\sqrt{x}-\sqrt{x}\ln (\sqrt{x})\] so that \[ \pi(x)\leq\frac{2\theta(x)}{\ln x}+\sqrt{x}\leq\left( 8\ln 2\right)\frac{x}{\ln x}+\frac{2x}{\ln x}=\left( 2+8\ln 2\right)\frac{ x }{\ln x}.\] We now prove Proposition 2. Since there are infinitely many prime numbers, we immediately find that $ \lim_{x\rightarrow\infty}\pi(x) =\infty$. Since $ \frac{1}{\ln x}\rightarrow 0$ as $ x\rightarrow\infty$, \textsc{Lemma 3.4} implies that \[ \lim_{x\rightarrow\infty}\frac{\pi(x)}{x}=0.\] Note We proved that $ \frac{\pi(x)}{\frac{ x }{\ln x}}\leq 2+8\ln 2$. In fact, it is known that \begin{thm} \textsc{Prime Number Theorem} We have \[ \pi(x)\sim\frac{ x }{\ln x}\] or \[ 1=\lim_{x\rightarrow\infty}\frac{\pi(x)}{\frac{ x }{\ln x}}.\] In 1896, Hadamard and Valle'e Poussin independently proved Prime Number Theorem using complex analysis. Several alternative proofs were known, including the elementary proofs of Paul Erd"os and Atle Selberg.

Peter wrote: Prove that there do not exist polynomials $ P$ and $ Q$ such that \[ \pi(x) = \frac {P(x)}{Q(x)} \] for all $ x\in\mathbb{N}$. suppose that $ k=Max\{degP,degQ\}$. we have $ \pi((k+2)!+2)=\pi((k+2)!+3)=...=\pi((k+2)!+k+2)=t$ therefore at $ k+1$ point we have $ P(x)-tQ(x)=0$ and $ deg(P-tQ)\leq Max\{degP,degQ\}=k$ therefore $ P-tQ$ is zero polynomial and for all $ x\in\mathbb{N}$ we have $ \pi(x)=\frac{P(x)}{Q(x)}=t$ but $ \pi(2)=1,\pi(3)=2$!contradiction!

ali666 wrote: Peter wrote: Prove that there do not exist polynomials $ P$ and $ Q$ such that \[ \pi(x) = \frac {P(x)}{Q(x)} \] for all $ x\in\mathbb{N}$. suppose that $ k = Max\{degP,degQ\}$. we have $ \pi((k + 2)! + 2) = \pi((k + 2)! + 3) = ... = \pi((k + 2)! + k + 2) = t$ therefore at $ k + 1$ point we have $ P(x) - tQ(x) = 0$ and $ deg(P - tQ)\leq Max\{degP,degQ\} = k$ therefore $ P - tQ$ is zero polynomial and for all $ x\in\mathbb{N}$ we have $ \pi(x) = \frac {P(x)}{Q(x)} = t$ but $ \pi(2) = 1,\pi(3) = 2$!contradiction! Nice, this is the way to go, but it is the same as s.tringali

For all positive integers $ n$, $ \pi(4n) = \pi(4n - 1)$. Hence it would follow that $ P(4n)Q(4n - 1) = P(4n - 1)Q(4n)$ for all positive integers $ n$. Since both sides are polynomials, it follows that $ P(x)Q(x - 1) = P(x - 1)Q(x)$ for all real numbers $ x$. In particular $ P(3)Q(2) = P(2)Q(3)$, so $ \pi(2) = P(2)/Q(2) = P(3)/Q(3) = \pi(3)$.