Proving that this function has the same value for all integers $\geq4$. My teacher gave us this question:
A function $f$ has the property
$$f(x+y) = f(xy)   $$
$$\forall x, y\geq4; x,y\epsilon Z$$
$$f(8)=9$$
Find $f(9)$.
I know the solution to this question.
$9=f(8) = f(4+4) = f(16)=f(8+8)=f(64)=f(16\cdot4)=f(20)=f(4\cdot5)=f(9)=9$
But I tried this for other values and everytime it came $9$. I think this will be true for every integer $\geq 4$. Intuition tells me that I can reach to any such integer from any such other integer by performing those operations. 
Is the function $9$ for every integer $\geq 4$? If so, what would be its rigorous proof? I think it may need some number theory.
Edit: I realized that if my assumption is true, it is only possible for $n\geq8$.
 A: There is no way to prove $f(5)=9$. 
Say $x+y=5$ and $x<y$ then $x<4$ and you can't use $f(5)=f(xy)$.
Say $xy=5$ then $x=1$ and $y=5$ (or vice versa) and now you can't use $f(5)=f(6)$. 
So we are stuck with $f(5)$. We have the same problem with $f(6)$ and $f(7)$.
Perhaps for all $n\geq 8$ we can say $f(n)=9$
A: We have no way to access $f(4)$, $f(5)$, $f(6)$, or $f(7)$ using only the information present.
(Due to time constraints, this is only a partial answer.  I invite anyone else to finish it or skewer it for hasty thinking.)
However, we can track our progress of understanding using points on the plane.  Let $n \geq 8$ and consider the line $y = n-x$ for $4 \leq x \leq n - 4$.  On this line, $f(x+y) = f(n)$, so the function is constant on each of these lines (but still potentially has different values on different lines).  (Note that we only speak of the points with integer coordinates on this line, but perhaps we draw a continuous line so we can more clearly visualize which points we have shown produce equal values in $f$.)
Now let $m \geq 16$ and consider the part of the hyperbola $y = \frac{m}{x}$ with $x \geq 4, y \geq 4$.  From $f(xy) = f(m)$, we see that $f$ is constant on each of these hyperbolae.
From these two facts, we see that each line, $L$, $f$ is constant and $f$ is forced to be the same constant value on every hyperbola intersecting $L$ at an integer point.  Symmetrically, each hyperbola forces the lines it integrally intersects to share its constant value for $f$.
These observations allow an inductive see-saw.  For any point, follow its line to the point with $y=4$.  Then follow that hyperbola to the point of minimum $|x - y|$.  This sequence of points at the end of the hyperbola move must move closer to the origin.  So we have a strictly decreasing bounded sequence of distances from the origin.  This means such a path from any point eventually lands on the $n=8$ line.  
(Last thought:  The case $y=4$, $m$ prime, makes me think we should actually take the minimum over all possible hyperbola moves from a line.  Maybe I'm being overcautious.)
