Problem 1704

Solution:

The only such function is $f(p, q)=p \cdot q$.
This function obviously 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 that the second condition is also satisfied.

It remains to 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 gives
$$p q+g(p, q)=1+\frac{1}{2}((p+1)(q-1)+g(p+1, q-1)+(p-1)(q+1)+g(p-1, q+1)),$$
that is
$$g(p, q)=\frac{1}{2}(g(p+1, q-1)+g(p-1, q+1)).$$
Hence 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 its outer terms $g(0, p+q)$ and $g(p+q, 0)$ are equal to zero, every term of the sequence has the value zero. Consequently $g(p, q)=0$ for all nonnegative integers, and therefore $f(p, q)=p \cdot q$ is the only solution.