The answer is $f(x) = x^2$ only, which clearly work. First, putting $m=n=1$ gives $f(1)=1$. Now, Putting $m=1$ gives $f(n) -n+1 \mid nf(n) + 1$, or equivalently, \[ f(n)-n+1 \mid n^2-n+1. \] Putting $m=n$ gives $2f(n)-n^2 \mid 2nf(n)$, or equivalently, \[ 2f(n) - n^2 \mid n^3. \] These equations are nice for philosophical reasons because they imply $f(n)$ has finitely many possible values for any $n$. Now, the $n^3$ is especially nice when $n$ is a prime, and in fact we claim that: Claim: For $p \ge 100$ a prime, we have $f(p) = p^2$. Proof. From $2f(p) - p^2 \mid p^3$, we have that \[ f(p) \in \left\{ \frac{-p^3+p^2}{2}, \frac{-p^2+p^2}{2}, \frac{-p+p^2}{2}, \frac{-1+p^2}{2}, \frac{1+p^2}{2}, \frac{p+p^2}{2}, \frac{p^2+p^2}{2}, \frac{p^3+p^2}{2} \right\}. \]Of these, the first two are negative hence impossible, and $f(p) = p^2$ is what we want. So we wish to exclude the other five cases. On the other hand, we know that \[ 2f(p) + 2p - 2 \mid 2p^2-2p+2 \]and so we can just manually check the five cases by hand (each in a routine way). $\blacksquare$ Once we have arbitrarily large primes, we are happy. Fix any $n$. Letting $m$ be a prime gives \[ p^2 + f(n) - pn \mid p^3 + nf(n) \iff p^2 + f(n) - pn \mid p^3 - np^2 + pn^2. \]By considering only $p > f(n)$ we may drop the factor of $p$ on the right, so \[ p^2 + f(n) - pn \mid p^2 - np + n^2 \iff p^2 + f(n) - pn \mid n^2 - f(n). \]By taking $p$ large, we conclude $f(n) = n^2$.

you have already solved these problems a long time ago)

This problem has emerged on AOPS like two years ago... very weird...

matol.kz wrote: you have already solved these problems a long time ago) Yes, I was an organizer for the American training camp (MOP), and I designed the mock exams for the camp using problems from the IMO Shortlist. This meant I had to solve (or try to solve ) all the problems myself.

I think when Dorlir Ahmeti says that his problem is difficulter than actual P3 he was speaking about N6.But isn't it too easy

Murad.Aghazade wrote: I think when Dorlir Ahmeti says that his problem is difficulter than actual P3 he was speaking about N6.But isn't it too easy Sorry I was mistaken, maybe I said that because I want it my problem to be P3, but is not easy problem.

This was also problem 2 at Bosnia Herzegovina TST 2017

Also German TSTST #3.

My solution is exactly same as that of v_Enhance, but posting it anyway.

As a sidenote, how this is N6 Clearly even the N3 is way harder than this.

Nice problem: I claim only $f(x)=x^2$ works, and this is easy to check. Suppose $f$ is a solution. Let $g(x)=f(x)-x^2$. Then, we have \[g(m)+g(n)+(m^2+n^2-mn)\mid mg(m)+ng(n)+m^3+n^3.\]Subtracting $m+n$ times the left from the right, we get \[g(m)+g(n)+(m^2+n^2-mn)\mid ng(m)+mg(n).\]Call this statement $P(m,n)$. We see that $P(n,n)$ implies $2g(n)+n^2\mid 2ng(n)$, so $2g(n)+n^2\mid n^3$, or $2f(n)-n^2\mid n^3$. Letting $n=1$, we see that $2f(1)-1\mid 1$, so $2f(1)=1\pm 1=0,2$. But $f(1)\ge 1$, so $f(1)=1$. Now, letting $n=p$ be prime, we see that \[2f(p)-p^2\in\pm\{1,p,p^2,p^3\}.\]Using the condition that $f(p)>0$, we get that \[f(p)\in\left\{\frac{p^2-p}{2},\frac{p^2-1}{2},\frac{p^2+1}{2},\frac{p^2+p}{2},p^2,\frac{p^2+p^3}{2}\right\}.\]Let's look at $P(p,1)$ now. We get that \[g(p)+(p^2-p+1)\mid g(p),\]so $f(p)-p+1\mid p^2-p+1$. Case 1: $f(p)=\frac{p^2-p}{2},\frac{p^2-1}{2},\frac{p^2+1}{2},\frac{p^2+p}{2}$. We'll just do the first part, the rest follow similarly. Here, we get that $p^2-p-2p+2\mid 2(p^2-p+1)$, so $p^2-3p+2\mid 2p^2-2p+1$, so $p^2-3p+2\mid 4p-3$. For large enough $p$ this is a contradiction. Case 2: $f(p)=\frac{p^2+p^3}{2}$. Here, the left side $f(p)-p+1$ is bigger than the right side for large enough $p$, so this is a contradiction. Thus, for large enough primes $p$, we get that $f(p)=p^2$, so $g(p)=0$. Now, $P(m,p)$ then implies \[g(m)+(m^2+p^2-mp)\mid p(m^2+p^2-mp).\]Suppose $g(m)\not=0$. We have that \[\frac{p(m^2+p^2-mp)}{g(m)+m^2+p^2-mp}=p-\frac{pg(m)}{g(m)+m^2+p^2-mp}\in\mathbb{Z}\setminus\{p\},\]so \[\frac{pg(m)}{g(m)+m^2+p^2-mp}\in\mathbb{Z}\setminus\{0\}.\]Taking $p$ to be very large, we see that the left side is on order $1/p$, so it eventually has norm less than $1$, a contradiction. Therefore, we must have $g(m)=0$, so $f(m)=m^2$ for all $m$, as desired. $\blacksquare$

The only solution is $f(n)=n^2$. Throughout the solution, denote $(x,y)$ as plugging $m=x,n=y$ into the functional equation. $(1,1) \longrightarrow 2f(1)-1 \mid 2f(1) \Longrightarrow f(1)=1$ Let $p \geq 5$ be an odd prime. $(p,p) \longrightarrow 2f(p) -p^2 \mid 2p^3$ Hence $f(p) \in \{ \frac{p^2 \pm 1}{2} , \frac{p^2 \pm p}{2}, p^2 , \frac{p^3+p^2}{2}\}$ $(p,1) \longrightarrow f(p) -p+1 \mid pf(p) +1 \Longleftrightarrow f(p) -p+1 \mid p^2 - p +1 \Longleftrightarrow f(p)-p+1 \mid p^2-2f(p) +p -1$ If $f(p) = \frac{p^2 \pm 1}{2}$ then $\frac{p^2 \pm 1}{2} -p+1 \mid p-1 \mp 1 \Longleftrightarrow p^2 \pm 1 -2p+2 \mid 2p-2\mp 2$ But $0 < 2p-2 \mp 2 +2p -2\leq 4p-2 < p^2 \pm 1 $ so not possible. If $f(p)=\frac{p^2 + p}{2}$ then $\frac{p^2 + p}{2} -p+1 \mid p^2 -p+1 \Longleftrightarrow p^2-p+2 \mid p^2 - p$ impossible If $f(p) =\frac{p^2-p}{2}$ then $p^2 -3p +2 \mid 4p$ again not possible. If $f(p)= \frac{p^3+p^2}{2}$ then $\frac{p^3+p^2}{2}-p+1 \mid p^2 -p+1 \Longleftrightarrow p^3 +p^2 -2p+1 \mid 2p^2 -2p+2$ not possible. Hence $f(p)=p^2 \ \ \forall p\geq 5$ $(p,n) \longrightarrow p^2 +f(n) -np \mid p^3 +nf(n)$ $$p^3 +nf(n) \equiv p^3 +n(np-p^2) \equiv n^2p -pf(n) \mod (p^2+f(n)-np)$$Take $p$ sufficiently large such that $p^2> n^2p -pf(n)-f(n)+np$ such a prime exists since the LHS is a quadratic in $p$ while RHS is linear. Then $n^2 p -pf(n) < p^2 +f(n) -np$ so $f(n)=n^2 \ \ \forall n \in \mathbb{N}$. Done.

@v_enhance your proof is nice but I think that this might be a typo here v_enhance wrote wrote: Claim: For $p \ge 100$ a prime, we have $f(p) = p^2$. Why $p>100$?

@above It's just saying concretely 100 is sufficient instead of saying it holds for all large $p$.This is just to avoid edge cases like $p=2$ that unnecessarily complicates the discussion.

Yay solved a $6$! (though this one's not too tough though) First let's get over with the obvious substitutions- $P(m,m): 2f(m)-m^2 \mid 2mf(m) => 2f(m)-m^2 \mid m^3$. Putting $m=1$ gives you $2f(1)-1 \mid 1 => f(1)=1$. $P(m,1): f(m)-m+1 \mid m^2-m+1 -m(f(m)-m+1) =>f(m)-m+1 \mid m^2-m+1$(Corollary which'll be useful: $f(m) leq m^2$. Fine, we already have quite a bit of information, and it's clear that there's probably going to be no way to solve a "functional divisibility" purely through algebraic manipulation- well let's just put in primes then, eh? For $p$ prime: $P(p,p): 2f(p) -p^2 \mid p^3$. Hey this seems pretty useful- it means $2f(p)-p^2$ must be equal to one of $(+-1,+-p,+-p^2,+-p^3)$. Well, we've got nothing much else to do, and it pretty much seems like $f(x)=x^2$ has got to be the only solution, so let's just bash it out using $f(m)-m+1 \mid m^2-m+1$? (and I promise it gets easier as you got through the cases) Case 1: $2f(p)-p^2=1 =>f(p)=\frac{p^2+1}{2}$ Thus $\frac{p^2+1}{2} -p+1 \mid p^2-p+1 => p^2-2p+3 \mid p^2-p+1 =>p^2-2p+3 \mid p-2 \neq$ possible. Case 2: $2f(p)-p^2 =-1 => f(p)=\frac{p^2-1}2$ Thus $\frac{p^2-1}2 -p+1 \mid p^2-p+1 =>p^2-2p+1 \mid 2p^2-2p+1-2(p^2-2p+1) =>p^2-2p+1 \mid 2p-1 \neq$ possible for $p \geq 5$. Case 3: $2f(p)-p^2=p=>f(p)=\frac{p^2+p}2$ Thus $\frac{p^2+p}2 -p+1 \mid p^2-p+1=>p^2-p+2 \mid 2p^2-2p+2 =>p^2-p+2 \mid 2 \neq$ possible. Case 4: $2f(p)-p^2=-p =>f(p)=\frac{p^2-p}2$ Thus $\frac{p^2-p}{2} -p+1 \mid p^2-p+1 =>p^2-3p+2 \mid 2p^2-2p+2 =>p^2-3p+2 \mid 4p-2 \neq$ possible for $p \geq 7$. Case 5: $2f(p)-p^2=-p^2 =>f(p)=0 \neq$ possible. Case 6: $2f(p)-p^2 =+-p^3$ $2f(p)-p^2=-p^3=>f(p)=\frac{p^2-p^3}2 <0 \neq$ possible for $p \geq 2$, and $2f(p)-p^2=p^3 \neq$ possible as $f(p) \leq p^2$ for $p \geq 2$ Aha, so which case did I sneakily not include when listing out these cases which end up not working- well, of course $f(p)=p^2$, which works for all $p$. Thus we can conclude $f(p)=p^2$ for $p \geq 7$. Now the problem's pretty easy to finish off- for a fixed $m$, choose some prime $p$ such that $(p,m)=(p,f(m))=1$. $P(m,p): f(m)+p^2-mp \mid p^3+mf(m) =>f(m)+p^2-mp \mid p^3 +mf(m)-m(f(m)+p^2-mp)=p(p^2-mp+m^2)$. As $(p,f(m))=1 =>f(m)-mp+p^2 \mid p^2-mp+m^2=> f(m)-mp+p^2 \mid m^2-f(m)$. Notice that we can choose arbitrarily large $p$ such that $p^2-mp+f(m)$ becomes arbitrarily large while $m^2-f(m)$ stays fixed- thus $f(m)-m^2$ must equal $0$ $=>f(m)=m^2$. Thus the only solution is $f(m)=m^2$, and it obviously works, hence we're done.

We claim the only solution is $f(n)=n^2$. We first show that $f(p)=p^2$ for sufficiently large primes $p$. Indeed, if $P(m,n)$ is the assertion, then $P(p,p)$ gives $2f(p) - p^2 \mid p^3$. So $2f(p)-p^2 = -p, 1, p, p^2, p^3,p^3$. Now it is not hard to show that the only possible solution is $f(p)=p^2$ is the only one that works (all others fall due to size reasons), for sufficiently large $p$. Now, $P(m,p)$ gives $f(m) + p^2-mp \mid mf(m) + p^3$, so Euclid gives $f(m) + p^2 - mp \mid p(p^2-pm+m^2)$. We want to show $f(m) = m^2$. So either $f(m) + p^2-mp \mid p^2-pm+m^2$ or $p\mid f(m) + p^2-mp\mid p(p^2-mp+m^2)$. If $f(m)<p$, then $p\nmid f(m)+p^2-pm$, so the latter case is impossible. Therefore, $f(m) + p^2-mp \mid p^2-pm+m^2$. Now $f(m)+p^2-mp \mid m^2-f(m)$. Taking sufficiently large $p$, this inequality must fail, unless $f(m)-m^2=0$, so $f(m)=m^2$.

Solution(edit: similar to above solution) We claim that the only solution is $f(x)=x^2$. This clearly works since $m^2-mn+n^2|m^3+n^3=(m+n)(m^2-mn+n^2)$. Let $P(m,n)$ be the assertion that $f(m)+f(n)-mn|mf(m)+nf(n)$. $P(1,1)$ gives $f(1)=1$ since $2f(1)-1|2f(1)$. $P(m,m)$ gives $2f(m)-m^2|m^3$. $P(1,m)$ gives $f(m)-m+1|m^2-m+1$. Thus $f(m) \leq m^2$ for all $m$. Note that $2f(2)-4|8$ implies $m=3,4$ or $6$. $f(2)-2+1|3$ now gives $f(2)=4$. $f(4)-3|13$ gives $f(4)=4$ or $16$. But $f(2)+f(4)-8$ is nonzero so $f(4)$ must be $16$. $f(3)-2|7$ gives $f(3)=3$ or $9$. Now, $P(3,4)$ gives $f(3)=9$ from here. Now, we make the following claim: $f(p)=p^2$ for all primes $p$. Proof: $2f(p)-p^2|p^3$ and $f(p)$ being a positive integer with $f(p) \leq p^2$ gives that $f(p)$ can be one of $\frac{p^2 \pm 1}{2}, \frac{p^2 \pm p}{2}, p^2$. For all cases other than $f(p)=p^2$, using $1+f(p)-p|p^2-p+1$ would give $p \leq 3$(trivial algebraic manipulations). But $f(2)=4, f(3)=9$. Thus $f(p)=p^2$ for all $p$. Proved. Now, since $f(n) \leq n^2$, all sufficiently large primes are coprime to $f(n)$. Choose a prime $p$ such that $p>>n^2$. Thus $P(p,n)$ with algebraic manipulations gives $p^2-pn+f(n)|p(p^2-pn+n^2)$. Since clearly $p$ is coprime to $f(n)$ we get $p^2-pn+f(n)|p^2-pn+n^2$. Now, $p>>n^2$ and $f(n) \leq n^2$ gives $f(n)=n^2$ for all $n$. Proved

Wow this just felt so easy for an N6. Proof: We claim that the only solution is $f(n)=n^2 \forall n \in \mathbb{N}$. Firstly notice that $P(1,1)$ $\implies$ $$2f(1)-1 \mid 2f(1) \implies f(1)=1$$Now we state a really powerful claim. Claim: Let $p$ be a large prime $\geq 1000$ then $f(p)=p^2$. Proof: Notice that $P(p,p) \implies$ $$2f(p)-p^2 \mid 2pf(p) \implies 2f(p)-p^2 \mid p^3$$So using this and the condition $f(p)\in \mathbb{N}$ we get that $$f(p) \in \left \{ \frac{p^2-p}{2},\frac{p^2+p}{2},\frac{p^2-1}{2},\frac{p^2+1}{2},\frac{p^2+p^3}{2},p^2\right\} $$Now notice that $P(p,1)$ $\implies$ $$f(p)-p+1 \mid p^2-p+1$$Now checking all possibilities of $f(p)$ (lots of algebra) we get that only $f(p)=p^2$ fits. $\blacksquare$ From here the proof smoothly follows.Notice that $P(n,1) \implies$ $$f(n)+1-n \mid nf(n)+1 \implies f(n)-n+1 \mid n^2-n+1 \implies f(n) \leq n^2$$Now for a particular $n$ fix $p$ as very large and such that $(p,f(n))=1$ (this holds since $f(n) \leq n^2$ ) then notice that $P(n,p) \implies$ $$f(n)+p^2-np \mid nf(n)+p^3 \implies f(n)+p^2-np \mid p^3-np^2+n^2p \implies f(n)+p^2-np \mid p^2-np+n^2 \implies f(n)+p^2-np \mid n^2-f(n)$$but notice that as $p$ is very large we have that $f(n)+p^2-np$ is much greater than $n^2-f(n) \implies \boxed{f(n)=n^2 \forall n \in \mathbb{N}}$ $\blacksquare$

GeoMetrix wrote: Claim: Let $p$ be a prime then $f(p)=p^2$. Proof: Notice that $P(p,p) \implies$ $$2f(p)-p^2 \mid 2pf(p) \implies 2f(p)-p^2 \mid p^3$$So using this and the condition $f(p)\in \mathbb{N}$ we get that $$f(p) \in \left \{ \frac{p^2-p}{2},\frac{p^2+p}{2},\frac{p^2-1}{2},\frac{p^2+1}{2},\frac{p^2+p^3}{2},p^2\right\} $$Now notice that $P(p,1)$ $\implies$ $$f(p)-p+1 \mid p^2-p+1$$Now checking all possibilities of $f(p)$ (lots of algebra) we get that only $f(p)=p^2$ fits. $\blacksquare$ I think this only works for $p$ sufficiently large. For example, if $p=3$ then $2f(3)-9 \mid 27$ is not enough to conclude $f(3)=9$.

Oh yeah @above. Thanks fixed it.

Felt pretty standard yet instructive. The answer is $f(n) = n^2$. Plugging in $m = n$ yields $2f(n)-n^2 \mid 2nf(n)$ which means $$2f(n)-n^2 \mid n^3$$Plugging in $n = 1$ here yields $f(1) = 1$. Plugging in $m = 1$ into the original FE yields $1+f(n)-n \mid 1+nf(n)$ which can be rewritten as $$1+f(n)-n \mid 1-n+n^2$$ Now let $n = p$ in the first relation. It follows that $2f(p)-p^2 = \{-p^3,-p^2,-p,1,p,p^2,p^3\}$. To check each case, look at the second relation above. Plugging in $p = 2$ and checking annihilates most of the cases and we're only left with $f(p) = \frac{p^2-p}{2}$ or $f(p) = p^2$. The latter clearly works. Plug in $p = 5$ to the former and it doesn't work, so we can only have $f(p) = p^2$. Now in the original FE plug in $m = p$ to get $$p^2+f(n)-pn \mid p^3+nf(n)$$This can be rewritten as $$p^2+f(n)-pn \mid p^3+nf(n)-n(p^2+f(n)-pn) = p(p^2-pn+n^2)$$but clearly $p^2+f(n)-pn \nmid p$ for large enough $p$ so $$p^2+f(n)-pn \mid p^2-pn+n^2$$which means $$p^2+f(n)-pn \mid n^2-f(n)$$Taking a large enough $p$ forces $f(n) = n^2$ as desired.

these nt problems are getting boring af they literally use the same concept over and over, feels the same as length and angle chasing 3000 lines in a problem; either this means isl is boring or im improving faster than i can see harder/more interesting nt isl but its probably the former and less the latter The answer is $f(x) = x^2$, which obviously works. (1,1) gives f(1)=1, (1,n) gives $f(n) -n+1 \mid nf(n) + 1-n(f(n)-n+1)= f(n)-n+1 \mid n^2-n+1$, (n,n) gives $2f(n)-n^2 \mid 2nf(n)-n(2f(n)-n^2)=n^3$. We claim that for a prime p, we have $f(p) = p^2$. Indeed, $2f(p) - p^2 \mid p^3$, checking the possible values of f(p), either they're nonpositive or they don't satisfy $2f(p) - 2p + 2 \mid 2p^2-2p+2$ (too lazy to type up details rn). Now, (p,n) gives $p^2 + f(n) - pn \mid p^3 + nf(n)-n(p^2+f(n)-pn) =p^3 - np^2 + pn^2$; for sufficiently large $p > f(n)$ the p factor doesn't matter on the RHS, namely $p^2 + f(n) - pn \mid p^2 - np + n^2-(p^2+f(n)-pn)=n^2 - f(n)$, and increasing p s.t. LHS>RHS implies RHS=0, namely $f(n) = n^2$ for all n.

This one is hilariously simple for an N6 $P(1,1)$ easily gives $f(1) = 1$. Now consider $P(p, p) \implies 2f(p)-p^2 | 2p^2$ so $f(p)$ is either $p^2, \frac{p^2+p}{2}, \frac{p^2 + 1}{2}$. Plugging $$P(p, 1) \implies f(p) -p + 1 | p^2 - p + 1$$. Note that $p^2 - p + 1 >\frac{p^2+p}{2}, \frac{p^2 + 1}{2} > \frac{p^2-p+1}{3}$ which makes these impossible. Thus $f(p) = p^2$ Now consider arbitrarily large $p$ and $$P(p, n) \implies p^2 + f(n)-pn | p^3 + nf(n) \implies p^2 + f(n)-pn | pn^2 - pf(n)$$by the Euclidian algorithm. For large $p$, this means $pn^2 = pf(n)$ so $f(n)\equiv n^2$.

Way too standard. Denote the assertion with $P(m, n)$. It follow by $P(1, 1)$ that $2f(1) - 1 \mid 2f(1)$ and thus $f(1) = 1$. Claim: If $m$ is a sufficiently large prime, then $f(m) = m^2$. Proof. Thus, by $P(m, 1)$ it follows that $f(m) + 1 - m$ divides $mf(m) + 1$ and thus $f(m) - m + 1$ divides $m^2 - m + 1$. Furthermore, by $P(m, m)$ \[ 2f(m) - m^2 \mid m^3 \]so $f(m) \in \left\{\frac{m^2 + 1}{2}, \frac{m^2 - 1}{2}, \frac{m^2 + m}{2}, \frac{m^2 - m}{2}, m^2, 0, \frac{m^2 + m^3}{2}, \frac{m^2 -m^3}{2}\right\}$. The last two possibilites grow at a cubic rate, $0$ is evidently impossible, The remaining $5$ can be checked, and only $f(m) = m^2$ works. $\blacksquare$ Claim: $f(n) = n^2$ for all $n$. Proof. Thus, by $P(m, n)$ for such $m$ it follows that $m^2 + f(n) - mn$ divides $m^3 + nf(n)$, so it follows that it also divides \[ m^3 + nf(n) - (m^2 + f(n) - mn)(m + n) = mn^2 - mf(n) \]As such, since this grows linearly, it follows from a sufficiently large $m$ that $mn^2 - mf(n) = 0$, or $n^2 = f(n)$. $\blacksquare$

Oh wowww(Shukur!!!) I just Solved My first N6 $f(m)+f(n)-mn | mf(m)+nf(n)$ $P(1,1)$ $\implies$ $f(1)=1$ Let's take any Prime $p$ From $P(1,p)$ $p^2 \geq f(p)$ $P(p,p)$ $2f(p)-p^2 | 2pf(p)$ $2f(p)-p^2| p^3$ So that $f(p)= \frac{p^2+1}{2} ; \frac{p^2+p}{2} ; p^2$ Case1: $f(p)=p^2$ $P(m,p)$ $f(m)+p^2-pm | mf(m)+p^3$ $f(m)+p^2-pm |p(p^2-pm+m^2)$ $f(m)+p^2-pm | m^2-f(m)$ Let's take $p$ sufficiently large so $f(m)=m^2$ Case 2: $f(p)=\frac{p^2+p}{2}$ $P(1,p)$ $p^2-p+2 | 2p^2-4p+2$ $p^2-p+2 | 2p-2$ $p^2+4 \le 3p$ but with $AM-GM$ $p^2+4 \geq 4p$ so contradicition. Case 3:$f(p)=\frac{p^2+1}{2}$ $P(1,p)$ $p^2-2p+5 | 2p^2-2p+2$ $p^2-2p+5 |2p-8$ Also By $AM-GM$ that is contradicition. So $f(m)=m^2$ is unique solution.....

Setting $m=n=1$, we have $2f(1) - 1 \mid 1$, or $f(1) = 1$. Setting $m=n$ and $m=1$ respectively, we have the two relations \begin{align*} 2f(m) - m^2 &\mid m^3 \\ 1+f(n) - n &\mid n^2-n-1. \end{align*}Setting $m=p$ in the first relation yields $$f(p) \in \left\{\frac{p^2+p^3}2, p^2, \frac{p^2+p}2, \frac{p^2+1}2\right\}.$$On the other hand, $f(p) \leq p^2$ by the second relation. Now, for sufficiently large $p$, \begin{align*} \frac 12(p^2-p+2) &\nmid p^2-p-1 \\ \frac 12(p^2-2p+3) &\nmid p^2-p-1 \end{align*}which eliminates all options but $f(p) = p^2$. Now we can extend to all $n$. Note that we have $$p^2+f(n) - np \mid p^3 + nf(n),$$so we can in fact write $$p^2+f(n) - np \mid p^3+nf(n) - (p+n)(p^2+f(n)-np) = p(f(n) - n^2).$$By picking sufficiently large $p$, the LHS is relatively prime to $p$, and furthermore it exceeds $f(n) - n^2$. This shows $f(n) = n^2$.

The solution is $f(x) \equiv x^2$. $P(m,m) \implies 2f(m) - m^2 \mid 2mf(m) \equiv m^3$. Now we substitute $m=1$ in this to get $f(1) = 1$. Now $P(m,1) \implies f(m) + 1 - m \mid mf(m) + 1 \equiv m^2 - m + 1$. Then we get that $f(m) \le m^2 \implies 2f(m) - m^2 \le m^2$. Now in $2f(m) - m^2 \mid m^3$, pick $m = p$ where $p$ is an odd prime. Then we get $2f(p) - p^2 \mid p^3$. But from the bound we got earlier, we get that only $\left\{-p,1,p,p^2\right\}$ might work. Now testing each possible value for $f(p)$ that we might get from above by substituting it into $f(m) + 1 - m \mid m^2 - m + 1$, we get that only $f(p) = p$ works. Now we can also find $f(2)$ by setting $m=2$ and then comparing the values we get from $f(m) + 1 - m \mid m^2 - m + 1$ and $2f(m) - m^2 \mid m^3$. This gives us $f(2) = 2$. Thus we get that $f(p) = p$ for all prime $p$. Now fix some $k \in \mathbb Z$, and we are going to find out $f(k)$. Let $S$ be the set of primes that are co-prime with $f(k)$. Now $P(k,p) \implies f(k) + p^2 - kp \mid kf(k) + p^3 \equiv p^3 - kp^2 + k^2p = p(p^2 - kp + k^2)$. Now note that $\gcd(f(k) + p^2 -kp,p) = 1$. Then we can reduce the previous condition to $f(k) + p^2 - kp \mid p^2 - kp + k^2 \equiv k^2 - f(k)$. But then note that we can take sufficiently large $p$ to prove that $k^2 \equiv f(k) \implies f(k) = k^2$ and we are done.

hurrah my firs n6 Let $P(x, y)$ be the given assertion. $P(1, 1)$ gives $2f(1) - 1 \mid 2f(1) \implies f(1) = 1$. Take $p$ to be a prime, $P(p, 1)$: $$f(p) + 1 - p \mid pf(p) + 1 \implies f(p) + 1 - p \mid p^2 - p + 1$$This means $f(p) \leq p^2$, and since $p^2 - p + 1$ is odd, either $f(p) = p^2$, or $p^2 - p + 1 > 3f(p) + 3 - 3p \implies \frac{p^2 + 2p - 2}{3}$. $P(p, p)$ gives $2f(p) - p^2 \mid p^3$, however plugging in the inequality back in we find that it's cannot be a divisor of $p^3$ (as it must be $-p$ or $-p^2$). So, $f(p) = p^2$. $P(p, n)$ gives: $$p^2 + f(n) - pn \mid p^3 + nf(n) \implies p^2 + f(n) - pn \mid p^3 + nf(n) - n(p^2 + f(n) - pn) = p^3 - np^2 + pn^2$$For sufficiently large $p$, we have: $$p^2 + f(n) - pn \mid p(p^2 - np + n^2) \implies p^2 + f(n) - pn \mid p^2 - np + n^2 \implies p^2 + f(n) - pn \mid p^2 - np + n^2 - p^2 - f(n) + pn = n^2 - f(n)$$This means that $n^2 - f(n) = 0 \implies f(n) = n^2$, so we are done.

We claim our only solution is $\boxed{f(x)=x^2}$, which works. Denote the assertion as $A(m,n)$. First note $A(1,1)$ gives $f(1)=1$. Consider a prime $p$: $A(p,p)$ tells us $2f(p)-p^2 = \pm 1, \pm p, \pm p^2, \pm p^3$, as \[2f(p)-p^2 \mid (2pf(p)) - p(2f(p)-p^2) = p^3.\] Using $A(p,1)$, casework on our previous 8 values tells us we can only have $f(p)=p^2$ for large $p$, as \[f(p)-p+1 \mid (pf(p)+1) - p(f(p)-p+1) = p^2-p+1.\] Notice our condition can be rewritten as \[f(m)+f(n)-mn \mid (m+n)(f(m)+f(n)-mn) - (mf(m)+nf(n))\]\[f(m)+f(n)-mn \mid m(f(n)-n^2)+n(f(m)-m^2).\] Fix $n$ as any natural. Setting $m=p$ as an arbitrarily large prime, we require \[f(n)+p(1-n) \mid f(n)-n^2,\] and since we can force $|f(n)+p(n-1)| > |f(n)-n^2|$ using large $p$, we must have $f(n)=n^2$ for all $n$, as desired. $\blacksquare$

We claim the answer is $f(x)=x^2$.

gives that $f(p)=p^2$ for prime $p$. Now consider $m$ and a sufficiently large prime $p$ with $f(m)<p$. Then $p^2+f(m)-pm \mid p^3+mf(m) \stackrel{\text{EA}}\iff p^2+f(m)-pm \mid p^3-mp^2+pm^2$. However $f(m)<p$ implies that the LHS is not divisible by $p$ so we require $p^2+f(m)-pm \mid p^2-mp+m^2 \stackrel{\text{EA}}\iff p^2+f(m)-pm \mid f(m)-m^2$. However as $p$ grows the LHS grows without bound while the RHS is fixed, therefore the RHS is $0$ or $f(m)=m^2$ for all $m$, and we are finished. $\square$

First note that clearly $f(1)=1$. Then, consider a prime $p$. We then have $2f(p)-p^2 \mid 2pf(p) \implies 2f(p)-p^2\mid p^3$. This gives us seven cases for what $2f(p)-p^2$ is ($1,p,-p,p^2,-p^2.p^3,-p^3$). One can check all these cases and see that for sufficiently large primes $p$ we must have $f(p)=p^2$. Clearly, $f(n)+f(1)-n \mid nf(n)+f(1) \implies f(n)-n+1\mid n^2-n+1$. This gives us that $f(n)\leq n^2$. Thus, we consider an integer $n$ and an arbitrarily large prime $p$ ($p>n^2+n$). This gives us that \[p^2-pn+f(n) \mid p^3+nf(n) \implies p^2-pn+f(n) \mid pf(n)-pn^2\]with some manipulation. Now, notice that we must have \[p^2-pn+f(n)\leq pn^2-pf(n)\implies (p+1)f(n)\leq p(n^2+n-p)<0\]which is a clear contradiction which implies that we must have $pn^2-pf(n)=0$ which requires $f(n)=n^2$ for all $n$ and we are done.

Nice! We claim the only solution is $f(x) = x^2$, which satisfies the equation. Sketch of proof: $P(1, 1)$ \implies $f(1) = 1$. $P(p, p)$ for $p$ primes gives some values of $f(p)$, most of which can be checked to fail for p large, apart from $f(p) = p^2$. So indeed $f(p) = p^2$ for all $p$ large. Now consider $P(x, p)$ with $p > x^2$, simplifying we get $f(x)-xp+p^2 \mid x^2-f(x)$. But now $x^2-f(x)$ has infinitely many divisors, thus it must be zero, implying the result. QED

dangerousliri wrote: Denote by $\mathbb{N}$ the set of all positive integers. Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for all positive integers $m$ and $n$, the integer $f(m)+f(n)-mn$ is nonzero and divides $mf(m)+nf(n)$. Proposed by Dorlir Ahmeti, Albania I think Nt6 could have been more difficult.

MuradSafarli wrote: dangerousliri wrote: Denote by $\mathbb{N}$ the set of all positive integers. Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for all positive integers $m$ and $n$, the integer $f(m)+f(n)-mn$ is nonzero and divides $mf(m)+nf(n)$. Proposed by Dorlir Ahmeti, Albania I think Nt6 could have been more difficult. This problem is pretty easy (I would say this and 2005 N6 are easiest N6s in the 21st century).

Plugging in $m=n=1$ gives $(2f(1)-1)|(2f(1))$ so $f(1)=1$. Plugging in $n=1$ gives $(f(m)+1-m)|(mf(m)+1)$ so $(f(m)+1-m)|(1-m+m^2)$. This also tells us that $f(m)\le m^2$. Plugging in $m=n$ gives $(2f(m)-m^2)|(2mf(m))$ so $(2f(m)-m^2)|m^3$. This also tells us that $f(p)=p^2$ for prime $p$. This is because $f(p)=\frac{p^2\pm p}{2}$ would give $\frac{2+(-2\pm 1)p+p^2}{2}|(1-p+p^2)$, which is absurd. This also tells us that $(f(m)+p^2-mp)|(mf(m)+p^3)$ for all prime $p$. This simplifies into $(f(m)+p^2-mp)|(p^3-mp^2+m^2p)$. Note that $\frac{p^3-mp^2+m^2p}{f(m)+p^2-mp}=p+\frac{p(m^2-f(m))}{f(m)+p^2-mp}$. We pick $p>69420m^2$ to get $(f(m)+p^2-mp)|(m^2-f(m))$. This implies $f(m)=m^2$, because otherwise, the RHS is much smaller than the LHS. Therefore, the only solution is $\boxed{f(x)=x^2}$. This works because $(m^2+n^2-mn)(m+n)=m^3+n^3$.