@above $ -x^m$ isn’t a solution, because the problem statement requires $P(n)$ to be a positive integer whenever $n$ satisfies $\omega(n)>2023^2023$.

@2above I might be misunderstanding your solution, but are you sure you didn't solve the problem for $\tau$ instead of $\omega$? I don't see how this power of $2$ will appear otherwise.

I think this is true Let $P(x)=a_vx^v+...+a_1x+a_0$. If $a_0=0$ let $P(x)=x^aQ(X)$ .Then for every $x$ we have that $Q(x)$ has the same primes as $x$.Let $Q(x)=b_sx^s+..+b_1x+b_0$ with $b_0$ different from $0$ select a prime $p$ with $p$ does not divese $b_0$ but $p|Q(x)$ then $p|x$ so $p|b_0$ contradiction so $Q(x)=+-1$. If $a_0$ different from $0$. Let $p_1,..,p_m$ be different primes who doesn't device $a_0$ and let $p_i|P(x_i)$ then $p_i$ doesn't device $x_i$ so by Chinese remainder Theorym and Direclert Theorym we can find $prime=p=x_i(mod p_i)$ Now let $r_1,r_2,...,r_{2023^{2023}}$ be different primes with $r_i=1(modp_1*..*p_m)$. If we take now $n=p*r_1*...*r_{2023^{2023}}$ contradiction by taking large $m$ So all the solution are: $P(x)=x^m$,$P(x)=c$ with $\omega(c)<=2023^{2023}+1$ and $c>0$

Did anyone find a solution without Dirichlet's Theorem?

Similar to above but likely easier to parse. Let $N=2023^{2023}$. We claim either (a) there is an $m\in\mathbb{N}$ such that $P(x)=x^m$ for all $x$ or (b) there is a $c\in\mathbb{N}$ with $\omega(c)\le N+1$ such that $P(x)=c$ for all $x$. The case $P$ is constant is clear, hence assume $P$ is non-constant. Let $P(x)=\alpha x^m Q(x)$ with $Q(0)\ne 0$. If $Q$ is constant, then $Q$ must clearly be one: otherwise if $Q=c$ with $|c|>1$ then by choosing primes $p_1<\cdots<p_{N+1}$ and letting $n=\prod_{1\le i\le N+1}p_i$, we arrive at a contradiction. Now assume $Q$ is non-constant, set $Q(x) = \alpha_d x^d+\cdots+\alpha_1 x+\alpha_0$ with $\alpha_d,\alpha_0\ne 0$. It is well-known that the set tof prime divisors of $Q$ is infinite. Let $q_1,\dots,q_T$ be distinct primes larger than $|\alpha_0|$ and $T\gg N$. Then there exists $a_i\ne 0$, $1\le i\le T$, such that $q_i\mid Q(a_i)$. Now let $p_1,\dots,p_{N+1}$ be distinct primes such that $p_1\equiv a_i\pmod{q_i}$ for $1\le i\le T$ and $p_j\equiv 1\pmod{q_1\cdots q_T}$ for $2\le j\le N+1$. Clearly such primes exist due to CRT + Dirichlet. We then let $n=p_1\cdots p_{N+1}$, so that $\omega(n)=N+1$, whereas $P(n)\equiv P(a_i)\equiv 0\pmod{q_i}$ so that $\omega(P(n))\ge T\gg N$, a contradiction.

Basically the same solution as the two above.

The answer is $P(x)=x^n$ or $P(x)=c$ where $\omega(c)\leq 2023^{2023}$ which clearly work. It is trivial to see that the given solutions are the only constant ones, as well as the only solutions of the form $cx^n$, henceforth suppose that $P$ has some factor which is not a power of $x$ (or constant). Let $N$ be the product of the first $2023^{2023}+1$ primes. First we prove that $x \nmid P(x)$. Suppose otherwise and let $P(x)=x^kR(x)$ where $k>0$ and $x \nmid R(x)$, so by Bezout's Lemma there exists some integer $M$ such that $\gcd(n, R(n))\leq M$ for all $n \in \mathbb{Z}^+$. On the other hand, by Schur's Lemma there exist arbitrarily large prime divisors of $R(Nx)$ (viewing this as a polynomial in $x$), so by picking $x$ such that $R(Nx)$ has a prime divisor $p>\max\{M,N\}$, so $p \mid P(Nx)$ but not $p \mid (Nx)^k$, so $\omega(P(Nx))>\omega(Nx)$ since every prime divisor of Nx is still a prime divisor of $P(Nx)$. Now we consider $x \nmid P(x)$ only. Let $Q(x)=P(Nx)$ so $x \nmid Q(x)$. By Schur's theorem we can find $2023^{2023}+3$ primes $p_1,p_2,\ldots$ not dividing $P(0) \neq 0$ and integers $a_1,a_2,\ldots$ such that $x \equiv a_i \pmod{p_i} \implies p_i \mid P(x)$, so we actually have $p_i \nmid a_i$ for all $i$. By the Chinese Remainder Theorem and Dirichlet, we can find some prime $P$ satisfying $P \equiv a_i \pmod{p_i}$ for all $i$, in which case $\omega(P(NP)) \geq 2023^{2023}+2$, but $\omega(NP)=2023^{2023}+2$: contradiction. Therefore no other polynomials work. $\blacksquare$ Remark: Kind of cute but also not particularly interesting. To some extent the entire problem should be routine to an experienced contestant, since I feel like this idea is simply the first thing you think of.

Let $P(x)=a_mx^m+a_{m-1}x^{m-1}+\dots+a_0$. Since $P(n)$ needs to be a positive integer for arbitrarily large positive integers $n$, we must have that $a_m>0$, which also implies that $P\not\equiv 0$. If $m=0$, then $P{}$ is a constant polynomial; let $P(x)=c$. Since we must have $\omega(n)>2023^{2023}$ for the given inequality to apply, for $P{}$ to satisfy the given condition it is necessary and sufficient that $\omega(c)\leq 2023^{2023}+1$. Else, if $P(x)=a_mx^m, m\ge 1$, i.e., all coefficients except $a_m$ are zero, then by choosing $n$ satisfying the conditions and coprime to $a_m$, we get that we need to have $$\omega(n)\ge \omega(P(n))=\omega(a_mn^m)=\omega(a_m)+\omega(n^m)=\omega(a_m)+\omega(n) \rightarrow \omega(a_m)\le 0\rightarrow \omega(a_m)=0$$Hence, we must have that $a_m=1$, i.e. $P(x)=x^m$, which is easily seen to also be a solution to the problem. Now, suppose that there is some $m>t\ge 0$ such that $a_t\neq 0$ and take $t$ to be minimal. If $t\neq 0$, then $P(x)=x^tQ(x)$, where $Q(x)=a_mx^{m-t}+\dots+a_t$. We choose a large enough $n$ such that $Q(n)>1$ and $n{}$ is coprime to $a_t$, which means that $n{}$ is also coprime to $Q(n)$, and we have $\omega(n)\ge \omega(P(n))=\omega(n^t)+\omega(Q(n))\ge \omega(n)+1$, impossible. Thus, we must have $t=0$. To finish off the problem, we have the following lemma: Lemma: Let $k\in \mathbb{N}$ be fixed, and $P$ be a non-constant polynomial such that $P(0)\neq 0$. Then, there is some $n\in\mathbb{N}$ such that $\omega(n)=k$ and $\omega(P(n))$ is arbitrarily large. The lemma evidently finishes the problem, since we can set $k=2023^{2023}+1$ to see that all polynomials of such type cannot satisfy the given conditions.

**Note: Dirichletless

My solution : The easy cases are when P si constant or $deg P \ge1$ and P(0)=0 which yeld the constant polynomial satisfying the relation and $x^d$ for some d So suppose deg P>=1 and P(0) is not 0 From Schur there exists an infinite number of primes $p_i$ which divideas some $P(n_i)$ and $p_i$ and P(0) are coprime. Choose k large and let $C=2023^{2023}$ Let's consider the system of congruences $X \equiv n_1 \pmod{p_1}$ $X \equiv n_2 \pmod{p_2}$ ........... $X \equiv n_k \pmod{p_k}$ and by CRT we get that $X \equiv T \pmod{p_1...p_k}$ and it is easy to see that T and $p_1..p_k$ are coprime so by Dirichlet there exists an infinite number of primes Take $n = q_1^{\alpha_1}q_2^{\alpha_2}\cdots q_(C+1)^{\alpha_(C+1)}$ so $\omega(n)>C$ so \[\omega(n)\ge\omega(P(n)).\]($q_1$ etc are primes from Dirichlet) But then it is easy that if we let ${\alpha_1 + \alpha_2 + \dots + \alpha_k} \equiv 1 \pmod{(p_1-1)...(p_k-1)}$then P(n) is divisible by $p_i$ so $\omega(P(n))>=k$ which is a contradiction because by hyphotesis we get that C+1>=k but k is large. So the only polynomials are $x^d$ and c

Kinda cringe that this problem is either you know Dirichlet's theorem and u get 10 or you don't and get 0. I know Dirichlet is common knowledge but it still doesn't seem fair.

Complete_quadrilateral wrote: Kinda cringe that this problem is either you know Dirichlet's theorem and u get 10 or you don't and get 0. I know Dirichlet is common knowledge but it still doesn't seem fair. Completely false. I didn’t use Dirichlet (actually, no theorem at all) and got a 10.

A solution by bootstraping powers of a fixed number: Assume $P(0)\neq 0$ Let $M=2023^{2023}$. Select $q_{i}$ $i=1,...,M+1$ primes and $p$ a prime different from the aforementioned and pick $c$ such that: $c\equiv x_{p} \pmod p$ where $p\mid P(x_{p})$ ($x_{p}$ exists by Schur's theorem) $c\equiv 0 \pmod q_{i}$ By CRT there exists such c. Now we construct the following sequence: $a_{1}=1$, $a_{n+1}=a_{n}+\lambda\prod_{i=1}^{n} (p_{i}-1)$ for several primes and some $\lambda$ specified below (that satisfy: $p_{i}$ divides $P(c^{a_{n}})$: $p_{1}=p$. On the n-th step we select $\lambda$ and $p_{n}$ Note that $P(c^{a_{n}})\equiv P(c^{a_{n-1}})\equiv 0 \pmod p_{i}$ for all $i=1,2,...,n-1$. Claim: $P(c^{x+\lambda y})$ where $y=\prod_{i=1}^{n-1} (p_{i}-1)$, $x=a_{n-1}$ has infinitely many prime divisors when $\lambda$ runs through $N$. Proof: Assume that's not the case. Call a prime $q$ "scary" if and only if $q\mid P(c^{x+\lambda y})$ for some $\lambda$. There are finitely many scary primes. For each scary prime $q$ not dividing c, set $m_{q}=U_{q}(P(c^{x}))$ and set $\lambda \to \lambda \prod_{q scary}(\phi(q^{m_{q}+1}))=\lambda s$. Then $P(c^{x+\lambda sy})\equiv P(c^{x})\neq 0 \pmod q^{m_{q}+1}$ hence $U_{q}P(c^{x+\lambda sy})\leq m_{q}$, hence upper bounded. If $q\mid c$ then $q\mid P(c^{sth})\equiv c^{sth}+P(0)$, and when sth is large enough, $U_{q}(c^{sth})=U_{q}(P(0))$ hence upper bounded. Setting for brevity $Y=\prod_{q scary} (\phi(q^{m+1}))(\prod_{i=1}^{n-1} (p_{i}-1))$ it suffices to prove that $P(c^{x+\lambda Y})$ has infinitely many prime divisors, to reach a contradiction. So, all scary primes have upper bounded p-adic valuations when considering the subsequence of $\lambda$. Hence, since $P(c^{x+\lambda Y})$ grows without bound, it should have infinitely many primes dividing some of its terms and the lemma is proven. Select now a prime $P\mid P(c^{x+\lambda_{P} y})$ and set $a_{n}=a_{n-1}+\lambda_{P}\prod_{i=1}^{n-1} (p_{i}-1)$ and $p_{n}=P$. Note that $\omega(c^{a_{n}})$ stays constant and $\omega(P(c^{a_{n}}))$ grows without bound, hence we derive a contradiction. So the only solutions are $P(x)=x^{d}$ and $P(x)=c$.

oVlad wrote: Complete_quadrilateral wrote: Kinda cringe that this problem is either you know Dirichlet's theorem and u get 10 or you don't and get 0. I know Dirichlet is common knowledge but it still doesn't seem fair. Completely false. I didn’t use Dirichlet (actually, no theorem at all) and got a 10. True but at least for me, i would never be able to solve it if I didn't know Dirichlet and i solved it in less that a hour cus i knew it. It still is a huge disadvantage if you don't know it since most of the solutions (on aops and i think on competition as well) use it.

oVlad wrote: Complete_quadrilateral wrote: Kinda cringe that this problem is either you know Dirichlet's theorem and u get 10 or you don't and get 0. I know Dirichlet is common knowledge but it still doesn't seem fair. Completely false. I didn’t use Dirichlet (actually, no theorem at all) and got a 10. Also could u please post your solution I would like to see it.

Let $P(n)=n^kQ(n)$ where $Q(0)\neq 0$. Let $p$ be a product of $2023^{2023}+1$ primes relatively prime to $Q(0)$. Then, if $Q$ is nonconstant, then by Schur's Theorem, there exists infinitely many primes dividing $Q(n)$ for some $n$, so there exists $m$ such that $Q(m)$ is divisible by a product of $2023^{2024}$ primes $q_i$ relatively prime to $p$ and $Q(0)$ with $\gcd(m,q_i)=1$ and $m\mid p$. Let $c=q_1q_2\ldots q_{2023^{2024}}$. Then, by Dirichlet's Theorem, there exists infinitely many primes of the form $cx+\frac mp$, so $\omega(pcx+m)=2023^{2023}+2$ and $\omega(P(pcx+m))\geq 2023^{2024}$, contradiction. Therefore, $Q$ is constant. If $k\geq1$ and $Q(n)>1$, then if $\gcd(n,Q(n))=1$, then $\omega(P(n))\geq\omega(n)+1$, contradiction. Therefore, we get either $P(n)=C$ for some $C$ satisfying $\omega(C)\leq2023^{2023}$ or $P(n)=n^k$, which both work.

Change $P(x)$ to $f(x)$ and assume it is non constant (for constant polynomial any constant $c$ with $\omega(c) \le 2023^{2023}$ works. Suppose that $f(0) \neq 0$ and let $M$ be a sufficiently large natural. Step 1: Arbitrary primes dividing the polynomial. Start with some prime $p_1$ where $p_1 \mid f(x_1)$ and $\gcd(x_1,p_1)=1$. Choose some $p_2 \neq p_1$ where $p_2 \mid f(x_1 + kp_1)$ (treat this as a polynomial in $k$, by Dirichlet we can select $x_2 = x_1 + kp_1$ as a prime and by Schur we can guarantee a $p_2$) such that $\gcd(p_2, x_2)=1$, so $p_1p_2 \mid f(x_2)$, again $p_1p_2 \mid f(x_2 + kp_1p_2)$, repeat the process to get to a point where we have an $x$ such that $P=\displaystyle\prod_{i=1}^M p_i \mid f(x)$ and $\gcd(P,x)=1$. Step 2: Construct more primes till you get a contradiction. Choose primes $c_i \equiv 1\pmod P$ for $1\le i\le M$ and let $C=\displaystyle\prod_{i=1}^M c_i$. Note that $f(Cr) \equiv f(r)\pmod{p_i}$ and also note that the set of primes dividing $f(kP+x)$ is infinite by Schur (treating this as a polynomial in $x$ as we can construct different $x$ each time through the above process) there we can guarantee the existence of more primes (say $a,b$) such that $ab \mid f(kP+x)$, finally choose $r=kP+x$ as a prime, then $abP \mid f(Cr)$. \[M+1 = \omega(Cr) \ge \omega(f(Cr)) \ge M + 1 + 1\]a contradiction. Step 3: The finish. Write $f(x)=x^ng(x)$ where $g(0) \neq 0$, suppose that $g$ is non constant then we can apply Schur and pretty much repeat the entire thing above to get a contradiction. So finally $f(x)=mx^n$, if there is some prime $p$ dividing $m$ then easily we can get a contradiction by selecting $x$ to be some number with $\nu_p(x)=0$, thus the final solution set is $f(x)=x^n$ for $n\ge 0$.

Complete_quadrilateral wrote: Also could u please post your solution I would like to see it. Of course. Let $C=2023^{2023}+1$. The case when $P{}$ is constant is trivial, so we shall not discuss it. If $x\mid P(x)$ then write $P(x)=x^kQ(x)$ so that $x\nmid Q(x)$. Hence, when $\omega(n)\geqslant C$ we have \[\omega(n)\geqslant \omega(P(n))=\omega(n^kQ(n))\geqslant \omega(n),\]so we must have equality. That being said, if $\omega(n)\geqslant C$ all the prime factors of $Q(n)$ must divide $n{}$. Assuming that $Q(x)$ isn't the constant polynomial 1, we may take $n{}$ to be the product of $C$ large enough prime numbers, which do not divide $Q(0)$. Hence, $Q(n)>1$ is relatively prime to $n{}$, contradiction. So, in this case, we get the solution $P(x)=x^k$ with $k\geqslant 1$. If $x\nmid P(x)$ we take $n=\Pi\cdot q^k$ where $\Pi$ is the product of $C$ distinct prime numbers relatively prime to $P(0){}$ and $q$ is another prime number relatively prime to $P(0)$. It follows that $C+1\geqslant \omega(P(\Pi\cdot q^k))$ for all $k\geqslant 1$. To put it simply, $\omega(P(\Pi\cdot q^k))$ is bounded. Hence, we may take some $t\geqslant 1$ for which $\omega(P(\Pi\cdot q^t))$ is maximal and write \[P(\Pi\cdot q^t)=\prod_{i=1}^m{r_i}^{s_i}.\]Note that $q\neq r_i$ for all $i=1,\ldots,m$. Choose $M>\max(s_i)$ and for some large enough positive integer $B{}$ consider \[A=t+B\cdot\varphi\left(\prod_{i=1}^m {r_i}^M\right).\]Clearly, all of $r_1,\ldots,r_m$ divide $P(\Pi\cdot q^A)$. By the maximality of $\omega(P(\Pi\cdot q^t))$ it follows that only these prime numbers can divide $P(\Pi\cdot q^A)$. However, \[P(\Pi\cdot q^A)\equiv P(\Pi\cdot q^t)\bmod{r_i^M}\implies \nu_{r_i}(P(\Pi\cdot q^A))=s_i.\]Therefore, all of the exponents in the prime factorization of $P(\Pi\cdot q^A)$ are bounded: they are equal to $s_i$ for $r_i$ and 0 otherwise. Taking $B{}$ large enough makes $A{}$ large enough such that $|P(\Pi\cdot q^A)|$ tends to infinity, contradiction (recall that we assumed that $P{}$ is non-constant).

Solved with hints from L567. Let $ P(x) = x^d \cdot Q(x)$ where $Q(0) \neq 0$. Assume possible $Q$ is non-constant. Call $k = 2023^{2023}$. By Schur we can pick pairwise distinct primes $p_1, p_2, \dots , p_{k+2} \in S$ such that for each $p_i$ we have a corresponding $b_i$; satisfying whenever $x \equiv b_i$ (mod $p_i$), we have $p_i \mid Q(x)$. Pick $p_i$ coprime to $Q(0)$ so that $p_i \nmid Q(0)$. Let $M = p_1p_2 \dots p_{k+2}$. By CRT, $x \equiv b_i$ (mod $p_i$) for all $i=1, 2, \dots, k+2$, is equivalent to $x \equiv c$ (mod $M$) for some $c$. Note $c$ is coprime with $M$. By Dirichlet, there is a prime $q$ such that $ q \equiv c$ (mod $M$). Choose $q_1, q_2, \dots, q_k$ to be any pairwise distinct primes, and different from all the primes picked till now. Consider $n = q (q_1 q_2 \dots q_k)^{\phi(M)}$. Note $\omega(n) = k+1$. But $n \equiv c$ (mod $M$). Hence $n \equiv b_i$ (mod $p_i$) and hence $p_i \mid Q(n) \mid P(n)$. Hence, $\omega(P(n)) \geqslant k+2$. But this contradicts given condition. Hence, $Q$ must be constant polynomial, say $c$. If $d \geqslant 1$: by $P(n)$ positive integer condition, $c \in \mathbb{N}$. If prime $p \mid c$, pick $n$ with $k+1$ prime factors none of which is $p$, to get contradiction. Hence $c=1$. This gives $P(x) = x^d$ as solution, which works. If $d=0$, $P \equiv c$, and it is easy to see precisely all $c$ such that $c \in \mathbb{N}$ and $\omega(c ) \leqslant k$ work. Hence, we are done!

The only answer are $P(x)=c$ where $\omega(c)\le 2023^{2023}+1$ and $P(x)=x^n$ for positive integers $n$. It's easy to check that these polynomials satisfies the problem condition. Now, we prove that those are the only solutions. We're going to use this lemma: Lemma 1. For any nonconstant polynomial P with integer coefficients, the set of all primes dividing $P$ is infinite. Proof. It's well known that there's infinitely many primes. Assume otherwise, let $p_1,p_2,...,p_N$ be the complete set of primes dividing $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$. If $a_0=0$, we can $x$ as the product of $N+1$ primes to get a contradiction. If $a_0\neq 0$. Notice that $P(a_0x)=a_na_0^nx^n+a_{n-1}a_0^{n-1}x^{n-1}+...+a_1a_0x+a_0=a_0\left( a_na_0^{n-1}x^n+a_{n-1}a_0^{n-2}x^{n-1}+...+a_1x+1 \right)$. We take $x=p_1p_2...p_N$. Now, $P(a_0p_1p_2...p_N)$ must have new prime divisor, a contradiction. $\blacksquare$ Now, we let $L=2023^{2023}$. It's clear that the first coefficient of $P$ must be a positive integer. Note that $P(x)=c$ where $\omega(c)\le L+1$ always satisfies the problem condition. Now, assume that $P$ is non constant. We write $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$. Case 1. If $a_0=0$, then let $P(x)=x^kQ(x)$ where $k \ge 1$, and the last coefficient in $Q(x)$ is nonzero. (a) If $Q$ is constant, we must have $|Q(x)|=1$, which forces $P(x)=x^n$. Otherwise, we can select $n$ to be the product of any $L+1$ primes that are greater than $Q(0)$ to get a contradiction. (b) If $Q$ is not constant, we select $n$ to be the product of $L+1$ primes that are greater than $Q(0)$ to get a contradiction. Case 2. If $a_0\neq 0$, let $p_1,p_2,...$ be the prime divisors of $P$ that are relatively prime to $a_0$ (there are still infinitely many of them). Let $p_i\mid P(x_i)$ for all $i=1,2,...,L+1$. We now choose $x\equiv x_i \pmod{p_i}$ for all $i=1,2,...,L+2$ (this exists by CRT) to get $p_1p_2...p_{L+2}\mid P(x)$. Let the CRT implies $x\equiv S \pmod{p_1p_2...p_{L+2}}$. We now have $\gcd(S,p_i)=1$ for all $i=1,2,...,L+2$. Now we choose primes $q_1,q_2,...,q_{L}\equiv 1\pmod{p_1p_2...p_{L}}$ and $q_{L+1}\equiv S \pmod{p_1p_2...p_{L+2}}$ (it exists by Dirichlet's Theorem). We now have $q_1q_2...q_{L+1}\equiv S \pmod{p_1p_2...p_{L+2}}$. It means that the number $$A=\frac{q_1q_2...q_{L+1}-S}{p_1p_2...p_{L+2}}$$is an integer. Now, we take $n=Ap_1p_2...p_{L+2}+S=q_1q_2...q_{L+1}$. Clearly, $\omega(n)=L+1$, but $\omega(P(n))\ge L+2$ which gives us a contradiction. Hence, the only answer are $P(x)=c$ where $\omega(c)\le 2023^{2023}+1$ and $P(x)=x^n$ for positive integers $n$, and it's easy to check that these polynomials satisfies the problem condition.

I really enjoyed this problem and it was quite simple for P3. My solution is basically the same as the ones above.

This is similar to the others. The only polynomials satisfy the conditions are $P\equiv c$ where $\omega(c)\leq 2023^{2023}+1$ and $P(n)=n^k$ where $k\in Z^+$. Suppose that $P$ is non-constant. Case $1$: The constant term of $P$ is not $0$. Let $c$ be the constant term. Rewrite $P(n)=nQ(n)+c$. If $Q\equiv 0,$ then $P\equiv 0$ holds. Suppose that $Q\not \equiv 0$. By Schur's theorem, we can choose $q_i|P(r_i)$ for $l$ primes. Take $q\equiv r_i(mod \ q_i)$ which is possible by Drichlet and chinese remainder theorem. We have $q_1...q_l|P(n)$ since $q_i|P(q)-P(r_i)$. Take a prime $p\equiv 1(mod \ q_1...q_l)$. We see that $ptQ(pt)\equiv tQ(pt)\equiv tQ(t)\equiv -c(mod \ q_1...q_l)$. Thus, $q_1...q_l|ptQ(pt)+c$. We can increase the number of distinct prime divisors of $t$. So we can find $q_1...q_l|nQ(n)+c$ for any $l,\omega(n)$. Note that $\omega(t)$ is equal to $1$ in the beggining and it is increased by exactly one at each step. Choose $l>\omega(n)>2023^{2023}$ which gives a contradiction. Case $2$: The constant term of $P$ is $0$. Let $P(n)=n^kR(n)$ where $R(0)\neq 0$. We have $\omega(n)\geq \omega(P(n))=\omega(n^kR(n))\geq \omega(R(n))$. The constant term of $R$ is not $0$ so $R$ must be constant. Let $R\equiv r$. $P(n)=rn^k$ and $\omega(n)\geq \omega(rn^k)$. If $r\neq 1,$ then we can choose $(n,r)=1$ and in this case $\omega(n^kr)=\omega(n)+\omega(r)\geq \omega(n)+1>\omega(n)$ which results in a contradiction. Thus, $r=1$ and $P(n)=n^k$ for any $k$ positive integer.$\blacksquare$

This problem was magical: Let $C=2023^{2023}$. If $P$ is a constant, then we are done. Assume that: $P$ is non-constant. If $P(0)=0$, then let $P(x) = x^k Q(x)$ where $Q(0) \neq 0$. Then, if $Q$ is constant, then let $Q \equiv c$. Plugging $n=p_1 p_2 \cdots p_{C+1}$ such that some prime $p|C$ and $p \neq p_i$ for all i. Then: $\omega(n) = C+1$ but $\omega(P(n)) \ge C+2$. Therefore, contradiction. If $Q$ is non-constant, then: by Schur's theorem, we have: $Q$ to have infinitely many prime divisors. Let $q_1, \cdots, q_T$ denote such divisors where $T$ is sufficiently larger than $C$, with $p_i | Q(x_i)$ for some $x_i \in \mathbb Z$. Construct primes distinct primes $p_1, \cdots, p_{C+1}$: $p_1 \equiv p_2 \cdots \equiv p_C \equiv 1 \pmod q_i$ and $p_{C+1} \equiv x_i \pmod q_i$. We can construct such primes due to CRT and Dirichlet's theorem (on Arithmetic Progressions). Taking $n=p_1 p_2 \cdots p_{C+1}$ gives: $\omega(n)=C+1$ but $\omega(P(n)) \ge T >> C$ as $P(n) \equiv P(x_i) \equiv 0 \pmod q_i$ for all $1 \le i \le T$. Therefore, contradiction and we are done. Therefore, the only solutions are: $P \equiv t$ for some constant $t$ with $\omega(t) \le C$ and $P(x) = x^k$ for some $k \in \mathbb N$.