International Zhautykov Olympiad

Putting $x=0$ in the original equation $$f(4x+3y)=f(3x+y)+f(x+2y)$$ we get $$f(3y)=f(y)+f(2y)$$ Next, for $y=-2x$ we have $f(-2x)=f(x)+f(-3x)=f(x)+f(-x)+f(-2x)$ (in view of the previous equation). It follows that $$f(-x)=-f(x)$$ Now, let $x=2z-v, y=3v-z$ in the original equation. Then $$f(5z+5v)=f(5z)+f(5v)$$ for all $z, v \in \mathbb{Z}$. It follows immediately that $f(5t)=tf(5)$ for $t \in \mathbb{Z}$, or $f(x)=\frac{ax}{5}$ for any $x$ divisible by 5, where $f(5)=a$. Further, we claim that $$f(x)=bx$$ where $b=f(1)$, for all $x$ not divisible by 5. In view of the previous equation, it suffices to prove the claim for $x>0$. We use induction in $k$ where $x=5k+r, k \in \mathbb{Z}, 0<r<5$. For $x=1$ the claim is obvious. Putting $x=1, y=-1$ in the original equation gives $f(1)=f(2)+f(-1)$ whence $f(2)=f(1)-f(-1)=2f(1)=2b$. Then $f(3)=f(1)+f(2)=3b$ by the previous equation. Finally, with $x=1, y=0$ we get $f(4)=f(3)+f(1)=3b+b=4b$. Thus the induction base is verified. Now suppose the claim is true for $x<5k$. We have $f(5k+1)=f(4(2k-2)+3(3-k))=f(3(2k-2)+(3-k))+f((2k-2)+2(3-k))=f(5k-3)+f(4)=(5k-3)b+4b=(5k+1)b; f(5k+2)=f(4(2k-1)+3(2-k))=f(3(2k-1)+(2-k))+f((2k-1)+2(2-k))=f(5k-1)+f(3)=(5k-1)b+3b=(5k+2)b; f(5k+3)=f(4\cdot 2k+3(1-k))=f(3\cdot 2k+(1-k))+f(2k+2(1-k))=f(5k+1)+f(2)=(5k+1)b+2b=(5k+3)b; f(5k+4)=f(4(2k+1)+3(-k))=f(3(2k+1)+(-k))+f((2k+1)+2(-k))=f(5k+3)+f(1)=(5k+3)b+b=(5k+4)b$. Thus the claim is proved. It remains to check that the function $f(x)=\frac{ax}{5}$ for $x$ divisible by 5, $f(x)=bx$ for $x$ not divisible by 5 satisfies the original equation. It is sufficient to note that 5 either divides all the numbers $4x+3y, 3x+y, x+2y$ or does not divide any of these numbers.