Sorry , I completely solved another problem !! I solved $nP(n!)=|P(n)|!$ Thanks : dinoboy Edit : small typo

I don't understand how putting $n=2$ results in $P(2) = 3$. When I put in $n=2$ I get $P(2) = 1$ or $2$

Yes , you are right !!

We have $P(n)$ in N for all n in N. Then $P(x) \in Q[x]$ Let $P(x)=x$ and P(x)=1 or 2 all x be a root. Now we do with P(x) not equal x for all x. Denote: $P(x)=a_{k}x^k+...+a_{0}$ Then $a_{k}>0$ ( if not then $P(n!)<0$ for large n) Note: $\mid P(x) \mid >x$ for all x>2.(*) Suppose that (*) not true then have $x_{0}$ and $\mid P(x_{0}) \mid \leq x_{0}$then $\mid P(x_{0}!) \mid \leq x_{0}!$ Hence have infinity x with $\mid P(x) \mid \leq x_{0}$ Then $P(x) -x \leq 0$ with infinity x then $P(x)=x-a$ a<0 that is not true. $P(x) \in Q[x]$ then have const m such that $m.P(x) \in Z$ P(n)>n all n >3 then $n!\mid P(n)!=P(n!)$ all n then $a_{0}=0$ Let k is max positive integer such that $a_{l}=0$ all l<k $P(x)=x^{k}Q(x)$ then $(n!)^{k}Q(n!)^{k}=(n^{k}Q(n))!$ l If k>1 then $n^k>(k+1)n$ all n large And $(n!)^{k+1} \mid ((k+1)n)!$ then $n! \mid (Q(n!)^{k}$ with all n large that is $a_{k}=0$ contradition! If k=1 similar with k>1 then $Q(n)<2$ all n But P(x) not equal x then $Q(n) =0$ P(x)=0 Contradition And of all we have root P(x)=x or P(x)=c all x with $x\in${1,2}

This year's USAMO. More solutions there.

Here's my solution.I'd really appreciate it if someone would check it. First,checking $P(n)$ constant gives that $\boxed{P(n)=1}$,$\boxed{P(n)=2}$. Let $P(x)=a_dx^d+...+a_kx^k$,where all the coefficents are nonzero. Also,since $P(n!) > 0$ for every positive integer $n$, it follows that $a_d > 0$. Pick a prime $q$ such that $(q,a_k)=1$.Note that $v_q(P(q!))=k$.Now lets suppose that $d > k$ for a moment.This means that we can choose our prime number so that $P(q) > cq^k$,for some $c$ which will be determined later.But now we have that $v_q( \lvert (P(q) \rvert !) > k$ if we pick $c \geq 2$,a contradiction. This contradiction now implies that $d=k$,which means that our polynomial has the form $P(x)=ax^b$.Plugging in values like $x=1$,$x=2$ and keeping in mind that $b > 0$ we can infer that $ \boxed{P(n)=n}$

Here is a detailed solution, (quite similar to what dragon did above), and hopefully is bug-free. Let's first handle the case of constant polynomials. If, for some $c$, $P(n)=c$ for every $n$, we get, $c=|c|!$, hence $c=1$ or $c=2$. Now, suppose that $P(\cdot)$ is non-constant, and let $P(n)=\sum_{i=0}^k a_in^i$, with $a_k>0$. $\bullet$ We first claim that $a_0=0$. To prove this claim, note that, a non-constant polynomial is unbounded (a well-known fact). Fix a prime number $p$ such that $p \nmid a_0$, and take $n$ large enough that both $|P(n)|>p$ and $n>p$. The first choice is possible, due to unboundedness of $P(\cdot)$, and the second is possible, since, had there not been infinitely many $n$'s with $|P(n)|>p$, definining, $$ S\triangleq \{n\in\mathbb{N}:|P(n)|>p\} \implies |S|<\infty, $$we would had, $$ P(n) \leq \max\left\{p,\max\left\{P(k):k\in S\right\}\right\}, $$a contradiction. For this choice of $p$ and $n$, since $|P(n)|>p$, $$ p \mid |P(n)|! \mid P(n!)=a_k(n!)^k+a_{k-1}(n!)^{k-1}+\dots+a_1 n! + a_0 \implies p\mid a_0, $$contradicting with the fact that $p\mid a_0$. Hence, $a_0=0$. Also note that, $P(2)$ must therefore be $2$, as $n\mid P(n)$, as a consequence of $a_0=0$. This further gives us $a_1\neq 0$, since, $$ a_k 2^k + \dots + a_1 2 = 2 \implies a_k 2^{k-1}+\dots+a_2 2 + a_1=1, $$hence, $a_1$ must be odd. In particular, $a_1\neq 0$. $\bullet$The degree of $P(\cdot)$ is at most one. To prove this claim, suppose that, ${\rm deg}(P)\geq 2$, namely, $k\geq 2$. As $P(n)$ is unbounded, $a_k>0$ (since, $P(n!)>0$). Hence, the polynomial, $Q(n)=P(n)-2n$, must eventually be non-zero. Now, take $p$ to be a prime, satisfying: $(i)$ $p \neq a_1$. $(ii)$ $P(p)\geq 2p$, The second is possible, as $a_1\neq 0$, there are (infinitely many) primes, that does not divide $a_1$, and are arbitrarily large. The second is a consequence of the fact that for large enough $n$, $P(n)\geq 2n$, as noted above. For this prime number, we will show that we are done. Now, since $|P(p)|\geq 2p$, $$ |P(p)|! \geq (2p)! \implies p^2 \mid |P(p)|! \implies p^2 \mid P(p!). $$Now, $$ p^2 \mid P(p!)=p!(a_k(p!)^{k-1}+a_{k-1}(p!)^{k-2}+\dots+a_2 p! + a_1), $$which implies that, $$ p \mid a_k(p!)^{k-1}+a_{k-1}(p!)^{k-2}+\dots+a_2 p! + a_1 \implies p \mid a_1, $$a contradiction. Note that, this argument is similar to our previous argument, where we have deduced that $a_0=0$, and this is precisely where we have used the fact that $a_0\neq 0$, hence, such a prime can always be found. $\bullet$ Finally, we are left with the case, $P(n)=an$, for some constant $a$. Evaluating $P(2)=2a=2$, we get that $P(n)=n$ satisfies the given condition. Thus the answer is, $\boxed{P(n)=\{1,2,n\}}$.

See that if $P(x)$ is constant, then $\boxed{P(x)=1}$ or $\boxed{P(x)=2}$. Now let $P(x)=a_kx^k+a_{k-1}x^{k-1}+\cdots+a_0$ where $k\ge 1$ and $a_k>0$. We have $P(1)=|P(1)|!$, hence $P(1)\neq 0$. Then, for all primes $p$, we have $|P(p-2)|!=P((p-2)!)= a_k (p-2)!^k+a_{k-1}(p-2)!^{k-1}+\cdots+a_0\equiv a_k+a_{k-1}+\cdots+a_0=P(1)\pmod{p}$. Since $P(1)\neq 0$, for large primes $p$, we have $|P(p-2)|!\equiv P(1)\not\equiv 0\pmod{p}$. Hence, $|P(p-2)|<p$ for all large primes $p$. Thus, we need to have $k=1$ and $a_k=1$. This gives us $P(x)=x+a_0$. Since $P(1), P(2)\in \{1,2\}$, we find that $a_0=0$ and $\boxed{P(x)=x}$.

emregirgin35 wrote: Find all polynomials with integer coefficients such that for all positive integers $n$ satisfies $P(n!)=|P(n)|!$ $$P(n!)=|P(n)|!$$$p > |P(-1)|$ $n \implies p-1 \implies |P(p-1)|! = P((p-1)!) \equiv P(-1)$ (mod$p$) $$|P(p-1)| <p \implies d(P) \leq 1$$Remaining easy.

lazizbek42 wrote: emregirgin35 wrote: Find all polynomials with integer coefficients such that for all positive integers $n$ satisfies $P(n!)=|P(n)|!$ $$P(n!)=|P(n)|!$$$p > |P(-1)|$ $n \implies p-1 \implies |P(p-1)|! = P((p-1)!) \equiv P(-1)$ (mod$p$) $$|P(p-1)| <p \implies d(P) \leq 1$$Remaining easy. You do not consider the case $P(-1)=0$ and I think this is why the above solution looks at $P(p-2)$

Easy but very cute problem . . This problem only need this easy lemma : $$a-b | P(a)-P(b)$$ If we put $n=1,2$ it is easy to see $P(1) , P(2) \in \{1,2\}$ Case 1 : $P(2)=2$ : Case 1.1 : $P(1)=1$ : $5 | P(3!)-P(1) = |P(3)|!-1$ and $4 | P(3!)-P(2)=|P(3)|!-2$ . Then we can see : $$|P(3)|=P(3!)=P(6)=6$$ Also we can see : $P(6!) = 6!$ . $$P((6!)!) = (6!)!,...........$$ Now it is easy to see in this subcase $P(x)=x$ . Case 1.2 : $P(2)=2$ , $P(1)=2$ : It is same as case 1.1 . In the case our polynomials is $P(x)=2$ . Case 2 : $P(2)=1$ We can see $n!-2 | P(n!)-1$ .If $n \geq 2$ it is Easy to see $P(x)=c$ . But we know $P(2)=1$ . Now it is easy to see $P(x)=1$ . Now our solutions are $P(x) = 1$, $P(x) = 2$ and $P(x) = x$ . $\blacksquare$ .