Jalil_Huseynov wrote: $f(x_1)!+f(x_2)!+\cdots f(x_3)!$ There is a small typo I guess, this should be $f(x_1)!+f(x_2)!+\cdots +f(x_k)!$.

BarisKoyuncu wrote: Jalil_Huseynov wrote: $f(x_1)!+f(x_2)!+\cdots f(x_3)!$ There is a small typo I guess, this should be $f(x_1)!+f(x_2)!+\cdots +f(x_k)!$. Fixed,Thanks!

Pretty standard in my opinion, but good one nevertheless.

Very nice problem indeed! First of all taking all $x_i$'s equal we get $f(x)\geq x$ since $f(x)!\mathrel{\vdots} x!$ Let $f(1)=c$ and $p\in \mathbb{P}$ and $p>c$. Take $x_1=x_2=\cdots =x_{k-2}=p$ (No problem if $k=2$) , $x_{k-1}=p-1$ and $x_k=1$. $\implies$ $(k-2)\cdot f(p)!+f(p-1)!+c!\mathrel{\vdots} p!\cdot (k-2)+(p-1)!+1\mathrel{\vdots} p$ (By Wilson's theorem). Since $f(p)\geq p$ and $c<p$ ,$p\mid f(p)!$ but $p\nmid c!$ $\implies$ $p\nmid f(p-1)!$ $\implies$ $f(p-1)\leq p-1$ and togather $f(p-1)\geq p-1$ we get $f(p-1)=p-1$ for all primes bigger that $c$. Now take any natural number $n$ and prime $q$ such that $q-1>max(f(n),c)$. Let $x_1=x_2=\cdots =x_{k-1}=q-1$ and $x_k=n$. $\implies$ $f(n)!+(k-1)\cdot f(q-1)!\mathrel{\vdots} (k-1)\cdot (q-1)!+n!$. Since $f(q-1)=q-1$ we get $f(n)!-n!\mathrel{\vdots} (k-1)\cdot (q-1)!+n!$ and since $RHS>LHS$ we get $LHS=0$ or $f(n)!=n!$. So $f(n)=n$ for all $n\in \mathbb{N}$. We are done!

hansu wrote: Pretty standard in my opinion, but good one nevertheless.

$k$ is fixed

Yes, indeed. It was a notational abuse on my side.

realize it is enough to prove $f(a)=a$ for infinite $a$ of large size bcs then \[x! + lots ~ of ~ a! \mid f(x)! + lots ~ of ~ a! \implies x! + lots ~ of ~ a! \mid f(x)! - x! \implies f(x)=x\]by wilson thm $(p-1)! \equiv -1 \mod p$. taking all $x_i$ to be equal gives $f(x) \ge x$. If $k=2k'$ select $k'$ of the $x_i$ to be $p-1$ and remaining $k'$ to be $1$ and $p$ is super large \[p \mid k'(p-1)! + k' \mid k' f(p-1)! + k'f(1)!\]if $f(p-1) > p-1$ then $p \mid k'f(1)$ but $p$ is very large so not possible hence $f(p-1)=p-1$ for infinitely many $p$ and we are done by above. If $k = 2k' - 1$ then select $k'$ of the $x_i$ to be $p-1$ one of them to be $2$ and rest of the $k'-2$ to be $1$ and similar procedure finishes solving $k=2$ and $3$ gives the idea

k=2 , P(a,a) f(a)!=(a!)c k=q+1, P(a,1,1,......,1), q is a big prime ==> f(1)!=c ==>f(a)=a

Solved with starchan, mueller.25, AdhityaMV Take $x_1 = x_2 = \cdots = x_k = n$ to get that $f(n) \geqslant n$ Take $x_1 = 1, x_2 = p-1, x_3,x_4, \cdots, x_k \geqslant p$ for some prime $p$. By Wilson, we obtain that $p \mid f(1)! + f(p-1)!$, implying that for $p > f(1)$, $f(p-1) = p-1$. Now, take $x_2 = \cdots = x_k = p-1$ for a big enough prime $p$ and fix $x_1 = x$. Then we get $x! + (k-1)(p-1)! \mid f(x)! + (k-1)(p-1)!$ which gives $x! + (k-1)(p-1)! \mid f(x)! - x!$. Since the right side is bounded and LHS can be made as large as needed, this forces that $f(x) = x$ for all $x$, which indeed works, so we're done. $\blacksquare$

Slightly different solution. In general, setting $x_i=x$ gives $$x!\mid f(x)!\implies x\leq f(x)$$ If $k=2$, $a!+b!\mid f(a)!+f(b)!$, take $a=1$ and $b=p-1$. By Wilson's Theorem, $p\mid 1+(p-1)!\mid f(1)!+f(p-1)!$. if $f(p-1)>p-1$ for some prime $p>f(1)$, then $$p\mid f(p-1)!\implies p\nmid f(1)!+f(p-1)!$$a contradiction. Thus, $f(p-1)=p-1$ for all prime $p>f(1)$. Then taking $b=p-1$ gives $$a!+(p-1)!\mid f(a)!+(p-1)!\implies a!+(p-1)!\mid f(a)!-a!$$so taking $p$ large gives $f(a)!-a!=0$ so $f(n)=n$ for all $n$. If $k>2$, take $x_{i+1}=f(x_i)$ for $i=1,2,\cdots, k-2$. so $x_1!+\cdots+x_k!\mid x_2!+\cdots+x_{k-1}! + f(x_{k-1})!+f(x_k)!$ giving $$x_1!+x_2!+\cdots+x_k!\mid f(x_k)!+f^{k-1}(x_1)!-x_k!-x_1!$$But by taking large $x_2$ and $2\neq 1,k$, we have that RHS is $0$ so since $x\leq f(x)\leq \cdots\leq f^{k-1}(x)$ we have $f(n)=n$ for all $n$.

Posting for Storage. Nice Problem! We show that $f(n)=n$ for all $n\in\mathbb{N}$, which indeed fits. Note that setting $x_1=x_2=n$ gives $n!\mid f(n)!\Longrightarrow f(n)\geq n$ Claim. For all primes $p$, we have $f(p-1)=p-1$. Proof. Consider $x_1!+x_2!\mid f(x_1)!+f(x_2)!$ and let $p>f(1)$ be a prime. Plugging $x_1=1$ and $x_2=p-1$, we have $(p-1)!+1\mid f(p-1)!+f(1)!$. However, from Wilson's Theorem, we have $p\mid(p-1)!+1$. So, $p\mid f(p-1)!+f(1)!$. Since $p>f(1)$, $p\nmid f(1)!\Longrightarrow p\nmid f(p-1)!\Longrightarrow f(p-1)<p$ This along with $f(p-1)\geq p-1$ gives $f(p-1)=p-1$. Also, $p\mid (p-1)!+f(1)!$ and $p\mid (p-1)!+1$ imply that $p\mid f(1)!-1$. If $f(1)>1$, then $f(1)!-1$ has infinitely many prime factors, which is impossible. So, $f(1)=1$. The above paragraph proves the claim for all $p>2$, while for $p=2$ we have $f(p-1)=f(1)=p-1=1$. $\square$ We now substitute $x_1=n$ and $x_2=p-1$ to get $n!+(p-1)!\mid f(n)!+(p-1)!\Longrightarrow n!+(p-1)!\mid f(n)!-n!$. Suppose $f(n)>n$. Then, $f(n)!-n!$ has an infinite number of factors that increase arbitrarily, which is impossible. So, $f(n)=n$ for all $n\in\mathbb{N}$, as claimed. $\square$

prime trick kills it