Baltic Way 1994 · Problem 7

Solution:

The sum has an even number of terms; they can be joined in pairs in such a way that the sum is the sum of the terms

$$\frac{1}{k^{3}} + \frac{1}{(p-k)^{3}} = \frac{p^{3} - 3 p^{2} k + 3 p k^{2}}{k^{3} (p-k)^{3}}.$$

The sum of all terms of this type has a denominator in which every prime factor is less than $p$ while the numerator has $p$ as a factor.