This problem has been proposed in Skhlarsky-Chentsov-Yaglom books, I have seen the proof for p=13 but I don't remember it, maybe I'll check it today. It's analog to the proof of any p prime. I remember the integer and rational one too, but never seen a proof for the real numbers. If for only integers, say that if $a_{1},a_2 ...a_p$ is a solution, then $ 0, a_2 - a_1 ... a_p - a_1$ is also a good solution. Also, if for some number q divides all of $a_{1},a_2 ...a_p$ then divide them all by p, look at the new sentence. This new sentence must contain an odd number, because if not then all numbers are divisible by 2 (Unless all numbers are 0). Now we have that this sequence at least one odd and one even (we have 0) number. Take an odd number x and even one y. Let S be the other p-2 numbers' sum. If S is odd, then taking away x, we'll have that S+y is also an odd, but it must be even for our statement to be true. If S is even, take away y. This completes the proof. The proof for rationals is also the same. In first place, you multiply them by a number r such that they become integers. It seems really hard for reals. I guess the problem was proposed wrongly

This is a less known method, but a really lovely one. This method is the same that solves many problems, for e.g. the 1999 Kürschák competition, problem 3.

The real case follows from the rational one from the existence of Hamel bases of R over Q(just look at the rational components of the terms in the basis that appear in the expansion of your original numbers).

It was very curious to me when bumped for the first time with this method. It is unusual for such kind of problems, since it is a pure elementary problem - some finite number of reals are given, which satisfy a simple property, and we want to prove they are all equal. It is not connected with set theory, we are not trying to find some set of reals with very pathological properties. I covered in two blog posts two Olympiad problems solvable by this method. It also can be applied in the following, more general, circumstances. Suppose we must prove that some property $ P$ about a set $ A$ of reals implies another property $ Q$ of the same set. Suppose both properties $ P$ and $ Q$ involve only linear expressions over elements of $ A$ that should be satisfied. Then, there is pretty good chance the described method to do the job.