[Peter]

Show that for all primes ,

[darij grinberg]

Peter wrote:

Show that for all primes ,

This is also a problem from the German National Olympiad (DeMO, 4th round) 2002, and I have written a solution for the op02 project. Since op02 does not seem to have survived, I am publishing the solution here:

We start by showing something really trivial:

Lemma 1. For any integer and any non-integer real number , we have .

Proof. Since is not an integer, we have for some real with . Hence, . Hereby, ; thus, , what yields Lemma 1.

For every such that , define the number . This number is trivially non-integer, since is not divisible by because is a prime and . Also, set . Then, Lemma 1 yields

.

This simplifies to

.

Now, the problem is straightforward:




,

so that

,

completing the solution.

darij

[ideahitme]

Peter wrote:

Show that for all primes ,

The same idea, but with different style.

Lemma. Let with . Then, we have



Proof. In case when , we also have . Then, we obtain and . Hence, . Now, we consider the case when . Since , we also have . Since is not an integer, one may write , where and . Observer that . Since and , this means that . It follows that


Lemma. Whenever , we obtain


Proof. Let . Since is prime, we see that is not divisible by so that is not an integer. Furthermore, from the congruence , we see that

The above proposition yields the desired result.


Now, we compute the summation. By symmetry, we obtain

and

Applying the above results we proved, we compute