Really? Just let $k=a+1$. Then it becomes equivalent to proving that $a^2$ divides $(a+1)^a-1$. The latter is $a(a)+1-1=0$ mod $a^2$ by the binomial theorem, so we're done. Finland has an easy olympiad...

yugrey wrote: Finland has an easy olympiad... Yes. I think that the mathematics in Finnish schools should be far more advanced. Still many students have difficulties to solve competition problems.

Well, that's what you get from an egalitarian educational system focused on getting everyone above a standard and not on cultivating talent. The Finnish system is world renowned because the test averages are high. However, the United States has a far better system. Although our average scores may be low, we have many excellent universities and we do well at International Olympiads (although the large population helps).

Fortunately I was able to spend my high school and college at the mathematically oriented class. But I was still unable to take part in international competitions. Mainly because the extra mathematics was more aimed to solve problems appeared in applied mathematics and industrial rather than competition problems. But every now and then there are Finnish scientists who has made some strong results.

Let's not turn this thread into a discussion of educational systems, especially using flawed comparisons by looking at a very small subset of people.

It should definitely be moved to pre olympiads or something. It's way too easy for this forum. The same holds with most of the other Finnish olympiad problems.