Olympiad Mathematics Again

Iberoamerican integer system of equations, 2018

June 30, 2019

Problem. For each integer n2, find all integer solutions of the following system of equations: x1=(x2+x3+x4++xn)2018, x2=(x1+x3+x4++xn)2018, xn=(x1+x2+x3++xn1)2018.

Origin: Iberoamerican Mathematical Olympiad 2018, problem 1.

Solution. First, notice that y20180 for any y, thus the right sides of all equations are non-negative. Therefore, so are the left sides, and so all xi0, 1in.

Now, let's consider xi0 which is the smallest among x1, \ldots, xn. Then we have xi0=(ii0xi)2018((n1)xi0)2018=(n1)2018xi02018. This can only be true either when xi0=0, or when xi0=1 and n1=1. Consider these two cases separately.

  • xi0=0. Then (ii0xi)2018=0. Then ii0xi=0. Then all xi=0. Finally, we can check directly that this result satisfies the original equations.
  • xi0=1 and n1=1. Then n=2. Without loss of generality, let's assume that i0=1. Then we have x1=1 and the two equations are 1=x22018 and x2=12018. Therefore, x2=1. Finally, we can check directly that this result satisfies the original equations.

To summarize, for n=2, there are two solutions: x1=x2=0 and x1=x2=1; for n3, there is only one solution: xi=0 for 1in.