It is interesting maybe two or three days ago I was trying to do a problem with identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ and I was trying to do a functional and it was exactly this form it is posted that I did wrote on paper. I was thinking when I have a more free time I would try to solve, but certanily someone was faster than me, it is a great problem. Also I think it should be not all the numbers $a,b,c$ equal to each other, otherwise $f(x)=x^2$ will not be a solution.

It's indeed nice and not hard. I have a solution using mostly putting $a$ or $a,b$ equal to $p+1$ which $p$ is a prime. I will post it later (due to lack of time near IMO)

dangerousliri wrote: Also I think it should be not all the numbers $a,b,c$ equal to each other, otherwise $f(x)=x^2$ will not be a solution. I think the convention is $0$ divides $0$, but not any other integers.

Also can someone tell me what IMOC stands for?

Nice problem. Here is the sketch of my solution: Step $1$: Do some boring casework with $f(1),f(2),f(3)$ to get $f(1)=1,f(2)=4$. Step $2$: Now use $P(a,1,1)$ and $P(a,2,2)$ to get $f(a)-2a+1|2(a-1)^2$ and $f(a)-4a+4|4(a-2)^2$. Step $3$: From the first condition, if $a=p+1$ for some prime $p$, then we see that $f(a) \in \{a^2,4a-3,2a+1,2a,3a-2,a,2a-2,1,2a^2-2a+1\}$. Step $4$: From the second divisibility condition, $f(a) \in \{a^2,4a-3\}$ if $p>61$, by using Euclidean algorithm. Step $5$: Use $P(a,a,1)$ now and use Euclidean algorithm to eliminate the $f(a)=4a-3$ case. Conclusion: For $p>61$ a prime, we have $f(p+1)=(p+1)^2$. Step $6$: Use $P(a,p+1,p+1)$ for large primes $p$ and use Euclidean algorithm to get $(f(a)-a^2)+(p+1-a)^2|2(p+1)(p+1-a)^2$. Thus for all large $p$, we have $(f(a)-a^2)+(p+1-a)^2|2(p+1)(a^2-f(a))$. If $f(a) \neq a^2$, the last part causes a contradiction when $p$ becomes sufficiently large in comparison to $a$. Thus, $f(a)=a^2$ which works clearly. Proved

@above nice solution Mathotsav. That's exactly my solutions too. [yey I'm able not to post mine anymore]

phoenixfire wrote: Also can someone tell me what IMOC stands for? IMOC is a non-official training camp in Taiwan mainly held by the past contestants in the summer vacation every year(mostly just after IMO, but not this year lol).

Mathotsav wrote: Nice problem. Here is the sketch of my solution: Step $1$: Do some boring casework with $f(1),f(2),f(3)$ to get $f(1)=1,f(2)=4$. Step $2$: Now use $P(a,1,1)$ and $P(a,2,2)$ to get $f(a)-2a+1|2(a-1)^2$ and $f(a)-4a+4|4(a-2)^2$. Step $3$: From the first condition, if $a=p+1$ for some prime $p$, then we see that $f(a) \in \{a^2,4a-3,2a+1,2a,3a-2,a,2a-2,1,2a^2-2a+1\}$. Step $4$: From the second divisibility condition, $f(a) \in \{a^2,4a-3\}$ if $p>61$, by using Euclidean algorithm. Step $5$: Use $P(a,a,1)$ now and use Euclidean algorithm to eliminate the $f(a)=4a-3$ case. Conclusion: For $p>61$ a prime, we have $f(p+1)=(p+1)^2$. Step $6$: Use $P(a,p+1,p+1)$ for large primes $p$ and use Euclidean algorithm to get $(f(a)-a^2)+(p+1-a)^2|2(p+1)(p+1-a)^2$. Thus for all large $p$, we have $(f(a)-a^2)+(p+1-a)^2|2(p+1)(a^2-f(a))$. If $f(a) \neq a^2$, the last part causes a contradiction when $p$ becomes sufficiently large in comparison to $a$. Thus, $f(a)=a^2$ which works clearly. Proved Nice one! ((the casework is indeed really boring lol my bad :p

phoenixfire wrote:

Anyone?

MarkBcc168 wrote: dangerousliri wrote: Also I think it should be not all the numbers $a,b,c$ equal to each other, otherwise $f(x)=x^2$ will not be a solution. I think the convention is $0$ divides $0$, but not any other integers. Then it should have been written as follows $$\forall_{(a,b,c)\in\mathbb{N}^3}\exists_{k\in\mathbb{Z}} \ af(a)+bf(b)+cf(c)-3abc=k\left(f(a)+f(b)+f(c)-ab-bc-ca\right).$$#1732

WolfusA wrote: MarkBcc168 wrote: dangerousliri wrote: Also I think it should be not all the numbers $a,b,c$ equal to each other, otherwise $f(x)=x^2$ will not be a solution. I think the convention is $0$ divides $0$, but not any other integers. Then it should have been written as follows $$\forall_{(a,b,c)\in\mathbb{N}^3}\exists_{k\in\mathbb{Z}} \ af(a)+bf(b)+cf(c)-3abc=k\left(f(a)+f(b)+f(c)-ab-bc-ca\right).$$#1732 What you have written is precisely the same as my description. Conventionally, $a$ is divisible by $b$ if and only if $a$ is a multiple of $b$.