From the equation we have $ \frac{P(a+b)+P(0)+P(c)}{a+b+c}$ is an integer . By the condition of problem we obtain : $ \frac{P(a+b)+P(c)+P(0)}{a+b+c}-\frac{P(a)+P(b)+P(c)}{a+b+c}$ is an integer for all integer $ a,b,c$ . Thus $ a+b+c|P(a+b)+P(0)-P(a)-P(b) ,\forall a,b,c\in Z$ ,take $ c$ c is large enough ,we must have $ P(a+b)+P(0)-P(a)-P(b)=0$ hence $ P(n)=kn+c$ forall n .Replace to the condition of problem ,it is easy to show that $ c=0$ and $ k$ is an integer . Thus all solutions are $ f(n)=kn,k\in Z$ .

TTsphn wrote: From the equation we have $ \frac {P(a + b) + P(0) + P(c)}{a + b + c}$ is an integer . By the condition of problem we obtain : $ \frac {P(a + b) + P(c) + P(0)}{a + b + c} - \frac {P(a) + P(b) + P(c)}{a + b + c}$ is an integer for all integer $ a,b,c$ . Thus $ a + b + c|P(a + b) + P(0) - P(a) - P(b) ,\forall a,b,c\in Z$ ,take $ c$ c is large enough ,we must have $ P(a + b) + P(0) - P(a) - P(b) = 0$ hence $ P(n) = kn + c$ forall n .Replace to the condition of problem ,it is easy to show that $ c = 0$ and $ k$ is an integer . Thus all solutions are $ f(n) = kn,k\in Z$ . You can't have $ P(kn)=kn$ from $ P(a+b)+P(0)-P(a)-P(b)=0$ Because $ P(n)$ uncertainty be real number.

math10 wrote: TTsphn wrote: From the equation we have $ \frac {P(a + b) + P(0) + P(c)}{a + b + c}$ is an integer . By the condition of problem we obtain : $ \frac {P(a + b) + P(c) + P(0)}{a + b + c} - \frac {P(a) + P(b) + P(c)}{a + b + c}$ is an integer for all integer $ a,b,c$ . Thus $ a + b + c|P(a + b) + P(0) - P(a) - P(b) ,\forall a,b,c\in Z$ ,take $ c$ c is large enough ,we must have $ P(a + b) + P(0) - P(a) - P(b) = 0$ hence $ P(n) = kn + c$ forall n .Replace to the condition of problem ,it is easy to show that $ c = 0$ and $ k$ is an integer . Thus all solutions are $ f(n) = kn,k\in Z$ . You can't have $ P(kn) = kn$ from $ P(a + b) + P(0) - P(a) - P(b) = 0$ Because $ P(n)$ uncertainty be real number. Yes, he can. $ P(a + b) + P(0) = P(a) + P(b)\Leftrightarrow P(a + b) - P(0) = (P(a) - P(0)) + (P(b) - P(0)),\forall a,b\in\mathbb{Z}$. Let $ Q(x): = P(x) - P(0)$ then $ Q(0) = 0$ and $ Q(a + b) = Q(a) + Q(b),\forall a,b\in\mathbb{Z}$. From that $ Q(n) = nQ(1),\forall n\in\mathbb{Z}$. Put $ k: = Q(1)$ then the two polynomials $ Q(n)$ and $ Q*(n) = kn$ are equal to each other in infinitely many points. Thus $ Q(n) = kn$, providing that $ P(n) = kn + c$ where $ c: = P(0)$.

Is it true ? Let $P(a,b,c) : a+b+c | p(a)+p(b)+p(c)$ . ( ) $P(a+b,0,c) : a+b+c | p(a+b)+p(0)+p(c)$ . ( ) If ( ) - ( ) we get $a+b+c | p(a+b)-p(a)-p(b)+p(0)$ . ( ) In ( ) if $\lim_{b \to +\infty }$ we get $p(a+b)=p(a)+p(b)-p(0) $ Then $p(x) =kx+c$ .

Note that from the condition we have that: \[ \frac{P(a)+P(b)+P(c)}{a+b+c} \in \mathbb Z \; \text{and} \; \frac{P(0)+P(b+c)+P(a)}{a+b+c} \in \mathbb Z \; \implies \frac{P(b+c)+P(a)+P(0)-P(a)-P(b)-P(c)}{a+b+c} \in \mathbb Z \]Now obviously the $P(a)$ is gone, but we still have $a$ in the denominator, now notice that for any integers $x,y,z$ satisfying $x+y+z \ne 0$ we have that $P(x)+P(y)+P(z)$ is an integer, therefore by using this twice we figure that $P(b+c)+P(0)-P(b)-P(c)$ is also an integer for any $b+c \ne -a$. Now we just pick $b,c$ positive integers and some really large $a$ so that the absolute value of the fraction above is $<1$ which would mean that it has to be zero, therefore we have $P(b+c)+P(0)=P(b)+P(c)$ for all $b,c>0$ integers, now of course you can fix $b$ and use that only the zero polynomial has infinite roots to get its true for all $c \in \mathbb C$ and similarily fix $b$ and do the same, and then repeat the process to get $P(b+c)+P(0)=P(b)+P(c)$ for all $b,c \in \mathbb C$ (call this $Q(b,c)$) now back to the main condition using $a=b=c$ we get that $P(a)$ is an integer for all integers $a \ne 0$, so from $Q(b,c)$ for $b,c \in \mathbb Z$ with $b+c \ne 0$ and $b,c \ne 0$ we get that $P(0)$ is an integer as well meaning that $P(a) \in \mathbb Z$ when $a \in \mathbb Z$, so consider $Q(b,c)$ for integers $b,c$ and let $P(x)=R(x)+P(0)$ then we have that $R(b+c)=R(b)+R(c)$, so from an easy induction we have that $R(n)=n \cdot R(1)$ for all $n$ integers, so from infinite roots being zero polynomial we have that $R(x)=d \cdot x$ for all $x \in \mathbb C$ meaning that $P(x)=dx+e$ for all $x \in \mathbb C$, now notice that we have $d, e \in \mathbb Z$ from previous observations, now replace to get that $e=0$ from size of $a+b+c$. Therefore only $P(x)=dx$ for $d \in \mathbb Z$ and for all $x \in \mathbb C$ work, thus we are done .