Let $P$ be the set of all ordered pairs $(p, q)$ of non-negative integers. Determine all functions $f: P \rightarrow \mathrm{IR}$ with the property

$$
f(p, q)=\left\{\begin{array}{c}
0 \quad \text { if } p q=0 \\
1+\frac{1}{2} f(p+1, q-1)+\frac{1}{2} f(p-1, q+1) \text { otherwise }
\end{array} .\right.
$$
The only such function is $f(p, q)=p \cdot q$.
This function clearly satisfies the first condition for $p=0$ or $q=0$. For $p q \neq 0$, we have $1+\frac{1}{2} f(p+1, q-1)+\frac{1}{2} f(p-1, q+1)=1+\frac{1}{2}(p+1)(q-1)+\frac{1}{2}(p-1)(q+1)=p q$, so the second condition is also satisfied.
Now it must be shown that there is no other function with these properties. To do this, we set $f(p, q)=p q+g(p, q)$. For $p q \neq 0$, the second condition yields $p q+g(p, q)=1+\frac{1}{2}((p+1)(q-1)+g(p+1, q-1)+(p-1)(q+1)+g(p-1, q+1))$, thus $g(p, q)=\frac{1}{2}(g(p+1, q-1)+g(p-1, q+1))$. Therefore, the numbers $g(0, p+q)$, $g(1, p+q-1), g(2, p+q-2), \ldots, g(p+q-1,1), g(p+q, 0)$ form an arithmetic sequence. Since their outer terms $g(0, p+q)$ and $g(p+q, 0)$ are both zero, each term in the sequence is zero. Consequently, $g(p, q)=0$ for all non-negative integers, and thus $f(p, q)=p \cdot q$ is the only solution.
Note: The second part of the proof is the more significant one. It is not enough to merely exclude certain types of functions, as some participants have attempted.