Shorth list for 2019?

wer wrote: Shorth list for 2019? No shortlist of last year 2018.

Let the assertion $n!+f(m)!\mid f(n)!+f(m!)$ be denoted by $P(m,n)$. Note that, $P(1,1)$ implies, $1+f(1)!\mid f(1)+f(1)!$, that is, $f(1)!+1\mid f(1)-1$. Consequently, we obtain that $f(1)=1$. Now, $P(1,n)$ yields $n!+1\mid f(n)!+1$, implying that $f(n)!\geq n!$ that is, $f(n)\geq n$. Now, fix $p$ to be an arbitrary prime. $P(1,p-1)$ yields, $(p-1)!+1\mid f(p-1)!+1$. Using Wilson's theorem, we have $p\mid (p-1)!+1$, and therefore, $p\mid f(p-1)!+1$. In particular, if $f(p-1)>p-1$, then $f(p-1)!$, being a product of at least $p$ consecutive positive integers, is divisible by $p$, and therefore, $f(p-1)!+1\equiv 1\pmod{p}$, a clear contradiction. Thus, $f(p-1)\leq p-1$. This, together with $f(n)\geq n$ for every $n$ yields $f(p-1)=p-1$, for every $p$ prime. Finally, $P(m,p-1)$ yields, $(p-1)!+f(m)!\mid f(m!)+(p-1)! \implies (p-1)!+f(m)! \mid |f(m!)-f(m)!|$. For any arbitrary $m$, by taking $p$ sufficiently large, we deduce, $f(m!)=f(m)!$. Using this one more time, we get $n!+f(m)!\mid f(n)!+f(m)!$, yielding $n!+f(m)!\mid |f(n)!-n!|$. Now, taking $m=p-1$ with $p$ sufficiently large, we get $f(n)!=n!$ for every $n$, yielding that $f(n)=n$, for every $n$.

grupyorum wrote: Let the assertion $n!+f(m)!\mid f(n)!+f(m!)$ be denoted by $P(m,n)$. Note that, $P(1,1)$ implies, $1+f(1)!\mid f(1)+f(1)!$, that is, $f(1)!+1\mid f(1)-1$. One of the factorials is supposed to be inside. So $RHS=f(1)+f(1!)$ Nvm I just don't know how to read.

@above Just see that $1+f(1)! |f(1)+f(1)! \implies 1+f(1)! | f(1)+f(1)!-f(1)!-1 \implies 1+f(1)! | f(1)-1$ It's Euclidean Algorithm.

Still didn't read properly

grupyorum wrote: Let the assertion $n!+f(m)!\mid f(n)!+f(m!)$ be denoted by $P(m,n)$. Note that, $P(1,1)$ implies, $1+f(1)!\mid f(1)+f(1)!$, that is, $f(1)!+1\mid f(1)-1$.

I'm dumb. Sorry 'bout that

Our TST this year.first P(1,1) implies that f(1) then claim that f(n)!>n!(or equal).Then P(1,p-1)for p a prime number.By using wilson and euclidean algorithm we get that f(p-1)=p-1.P(p-1,m) and you are almost done :)

Lordi wrote: Our TST this year.first P(1,1) implies that f(1) then claim that f(n)!>n!(or equal).Then P(1,p-1)for p a prime number.By using wilson and euclidean algorithm we get that f(p-1)=p-1.P(p-1,m) and you are almost done :) Can you tell me which contest and which problem it was I can put in the source.

dangerousliri wrote: Lordi wrote: Our TST this year.first P(1,1) implies that f(1) then claim that f(n)!>n!(or equal).Then P(1,p-1)for p a prime number.By using wilson and euclidean algorithm we get that f(p-1)=p-1.P(p-1,m) and you are almost done :) Can you tell me which contest and which problem it was I can put in the source. Albanian TST problem 1

My solution: $n!+f(m)!\vert f(n)!+f(m!)$ Substituting $m=n=1$ we get $f(1)+1\vert f(1)-1$. Therefore $f(1)=1$ Substituting $m=1$ in the given condition,we get $1+n!\vert f(n)!+1$ and substituting $n=1$,we get $1+f(m)!\vert 1+f(m!)$.Therefore $1+n!\vert 1+f(n)!\vert 1+f(n!)$.........(1) Now ,by Wilson's theorem, $p\vert (p-1)!+1$ where p is a prime Now from (1) we get that $p\vert f((p-1)!)-f(p-1)!$ Also by putting $m=n=p-1$,we get $(p-1)!+f(p-1)!\vert (p-1)!-f((p-1)!)$ Now here is the interesting part:$p$ divides the divisor but $p$ does not divide the dividend So we conclude that $f((p-1)!)=(p-1)!$ and consequently $f(p-1)=p-1$ from (1).Now fix $n$ and vary $m$ over $p-1$ such that $p$ is prime.So we get $f(n)=n$ for all natural numbers $\blacksquare$

Excellent problem! Here's my solution, pretty much the same as others

dangerousliri wrote: Find all functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $$n!+f(m)!|f(n)!+f(m!)$$for all $m,n\in\mathbb{N}$ Proposed by Valmir Krasniqi and Dorlir Ahmeti, Albania Let $P(m,n)$ the assertion of the given FE: $P(1,1)$ and assume that $f(1) \ne 1$ $$1+f(1)! \mid f(1)!+f(1) \implies 1+f(1)! \mid f(1)-1 \implies f(1) \ge 2+f(1)!$$Clearly a contradiction since for all $n \in \mathbb N$ this is truth $n! \ge n$. Now $f(1)=1$ . $P(1,n)$ $$n!+1 \mid f(n)!+1 \implies f(n)! \ge n! \implies f(n) \ge n$$$P(1,p-1)$ $(p \ge 3)$ and assume that $f(p-1)>p-1 \implies f(p-1)!$ has at least $p$ factors $$(p-1)!+1 \mid f(p-1)!+1 \implies p \mid f(p-1)!+1 \implies p \mid 1$$Now this is a contradiction and hence $f(p-1)=p-1$ for all primes $p \ge 3$. $P(n,p_{\infty}-1)$ where $p_{\infty}$ is a kind larger prime. $$(p_{\infty}-1)!+f(n)! \mid f(n!)+(p_{\infty}-1)! \implies (p_{\infty}-1)!+f(n)! \mid f(n!)-f(n)! \implies f(n)!=f(n!)$$$P(p_{\infty}-1,n)$ $$n!+(p_{\infty}-1)! \mid f(n)!+(p_{\infty}-1)! \implies n!+(p_{\infty}-1)! \mid f(n)!-n! \implies f(n)!=n! \implies f(n)=n$$Then the only solutions is: $\boxed{f(n)=n \; \forall n \in \mathbb N}$ Thus we are done

As usual, let $P(n,m)$ denote the assertion $n!+f(m)!\mid f(n)!+f(m!)$ and let $p$ denote an arbitrary prime, $p_{\infty}$ denote an arbitrarily large prime. $\textbf{1. }f(1)=1,f(n)\ge n$ $P(1,1)\Rightarrow 1+f(1)!\mid f(1)!+f(1)\Rightarrow 1+f(1)!\mid f(1)-1\Rightarrow f(1)=1$ $P(n,1)\Rightarrow n!+1\mid f(n)!+1\Rightarrow f(n)!\ge n!\Rightarrow f(n)\ge n$ $\textbf{2. }f(p-1)=p-1$ $P(p-1,1)\Rightarrow (p-1)!+1\mid f(p-1)!+1\Rightarrow p\mid f(p-1)!+1$ by Wilson's. If $f(p-1)>p-1$ then $p\mid f(p+1)!$, so $f(p+1)!+1\equiv1\pmod p$, contradiction. Then $f(p-1)\le p-1$, so $f(p-1)=p-1$ as desired. $\textbf{3. }f(n!)=f(n)!$ $P(p_{\infty}-1,n)\Rightarrow(p_{\infty}-1)!+f(n)!\mid(p_{\infty}-1)!+f(n!)\Rightarrow(p_{\infty}-1)!+f(n)!\mid f(n!)-f(n)!$, so we need $f(n!)-f(n)!=0$. $\textbf{4. }\text{finishing}$ $P(n,p_{\infty}-1)\Rightarrow n!+(p_{\infty}-1)!|f(n)!+f((p_{\infty}-1)!)\Rightarrow n!+(p_{\infty}-1)!|f(n)!+(p_{\infty}-1)!\Rightarrow n!+(p_{\infty}-1)!|f(n)!-n!$, so $f(n)!=n!$, hence $\boxed{f(n)=n}$.

Actually same but I'm going to post my solution anyway... Let $P(m,n)$ denote the assertion and $p$ arbitrary prime number $P(1,1) \implies 1+f(1)! | f(1)! + f(1) \implies 1+f(1)! | f(1)-1 \implies$ obviously $f(1)=1$ $P(1,p-1) \implies (p-1)!+1 | f(p-1)!+1(*)$ By Wilson $(p-1)!+1\equiv0\pmod p \implies f(p-1)!+1\equiv0\pmod p(**)$ By $(*)$ $f(p-1) \ge (p-1)$ , by $(**)$ $f(p-1)<p$ Thus $f(p-1)=p-1$ $P(m,p-1) \implies (p-1)!+f(m)! | (p-1)! + f(m!) \implies (p-1)!+f(m)! | f(m!)-f(m)!$ , We can always find greater and greater $p$ such that $LHS>RHS$ Thus $f(m!)-f(m)! = 0 \implies f(m!)=f(m)!$ $P(p-1,m) \implies (p-1)! + m! | f(m)! + f((p-1)!) \implies (p-1)! + m! | f(m)! + (p-1)! \implies (p-1)! + m! | f(m)!-m!$ with same idea we used before $f(m)=m$ so we are done..

$P(1,1)\Longrightarrow f(1)!+1\mid f(1)!+f(1)\Longrightarrow f(1)!+1\mid f(1)-1\Longrightarrow f(1)=1$ $P(p-1,1)\Longrightarrow (p-1)!+1\mid f(p-1)!+1\Longrightarrow f(p-1)!\geq (p-1)!$, but $p\mid (p-1)!+1$, so $p\mid f(p-1)!+1$, so $f(p-1)<p$ and so $f(p-1)=p-1$ $P(p-1,m)\Longrightarrow (p-1)!+f(m)!\mid (p-1)!+f(m!)\Longrightarrow (p-1)!+f(m)!\mid f(m!)-f(m)!\Longrightarrow f(m!)=f(m)!$ Now with the previous line, $P(n,p-1)\Longrightarrow n!+(p-1)!\mid f(n)!+(p-1)!\Longrightarrow n!+(p-1)!\mid f(n)!-n!\Longrightarrow f(n)=n$.

my solution is same with others but i will post for storage $n!+f(m)!| f(n)!+f(m!)$ $P(1,1)$ $\implies$ $f(1)=1$ let $p$ be prime number $P(p-1,1)$ $(p-1)!+1|f(p-1)!+1$ $f(p-1)! \geq (p-1)!$ $p|(p-1)!+1|f(p-1)!+1-(p-1)!-1$ if $f(p-1)! > (p-1)!$ $p|(p-1)!$ which is contradicition. $f(p-1)=p-1$ $P(1,p-1)$ $1+(p-1)! |1+f((p-1)!)$ with same idea which i metion above $f((p-1)!)=f(p-1)!=(p-1)!$ $P(p-1,m)$ $(p-1)!+f(n)! |(p-1)!+f(n!)$ $f(n)!=f(n!)$ $P(n,p-1)$ $n!+(p-1)!|f(n)!-n!$ take $p-1$ sufficiently large then $f(n)!=n!$ $f(n)=n$ is solution.

We claim the only solution is $\boxed{f\equiv \text{id}}$, which clearly works. We now prove it is the only one. Claim $f(1)=1$. Proof: Set $m=n=1$ to get $1+f(1)!\mid f(1)! + f(1)$. Thus, $\tfrac{f(1)!+f(1)}{f(1)!+1} = 1+\tfrac{f(1)-1}{f(1)!+1}$ is an integer. But $f(1)!+1>f(1)-1$, so $f(1)=1$, as desired. Claim: $f(n)\ge n$ for all $n$. Proof: Set $m=1$ to obtain $n!+1\mid f(n)!+1$. Clearly, $n!+1\le f(n)!+1$, which proves the desired. Claim: $f$ has arbitrarily large fixed points. Proof: Let $p$ be a prime, and set $n=p-1$, $m=1$. We must have $(p-1)! + 1\mid f(p-1)!+1$. By Wilson, $p\mid (p-1)!+1$, so $p \mid f(p-1)!+1$. If $f(p-1)>p-1$, then this would not be the case, so $f(p-1)=p-1$, which proves the claim. Claim: $f(m)!=f(m!)$ for all $m$. Proof: Set $n$ as a fixed point of $f$; then, $\tfrac{n!+f(m!)}{n!+f(m)!} = 1+ \tfrac{f(m!)-f(m)!}{n!+f(m)!}$ must be an integer. Then, choose an $n>f(m!)$, which we know we can do from our above claim, and thus $f(m!)=f(m)!$, as desired. Claim: $f(n)=n$ for all $n$. Proof: Using the previous claim, we must have $\tfrac{f(n)!+f(m)!}{n!+f(m)!} = 1+\tfrac{f(n)!-n!}{n!+f(m)!}$ be an integer. Then, simply pick an $m$ such that $f(m)>f(n)$, and now we are forced to have $f(n)!=n!$. Thus, $f(n)=n$, and we are done. $\blacksquare$

From $m=n=1$ we get $f(1)!+1\mid f(1)!+f(1)$, so $f(1)!+1\mid f(1)-1$ and hence $f(1)=1$. From here $m=1$ gives $n! + 1 \mid f(n)! + 1$ and so $f(n) \geq n$ for all $n$. Now let $p$ be an arbitrary prime number. By Wilson's theorem we have $p \mid (p-1)! + 1$, so setting $m=p-1$, $n = 1$ yields $(p-1)!+1\mid f(p-1)!+1$, so $p \mid f(p-1)! + 1$, which implies $f(p-1) < p$. Together with $f(p-1) \geq p-1$ from the first paragraph, we obtain $f(p-1)=p-1$ for any prime $p$. Now let $n=p-1$ for a prime $p$, then $(p-1)!+f(m)!\mid (p-1)!+f(m!)$, so $(p-1)!+f(m)!\mid f(m!)-f(m)!$ and large $p$ imply $f(m!)=f(m)!$. Substituting back in the initial problem and then setting $m=p-1$ and $n$ to be arbitrary yields $n!+(p-1)!\mid f(n)!+(p-1)!$, so $n!+(p-1)!\mid f(n)!-n!$ and taking large $p$ concludes $f(n) = n$.

$\boxed{ANSWER:f(n)=n}$ is my solution wrong?

Sorry for wrong solution, this is correct