The prime descomposition is 2010 = 2 * 3 * 5 * 67, so by Fermat's little theorem and the chinese remainder theorem any (inversible) element x modulo 2010 (not a mutiple of 2, 3, 5 or 67) will satisfy x¹³² = x^{mcm(2 - 1, 3 - 1, 5 - 1, 67 - 1)} = 1. It just so happens that 2¹² = 4 + 132 * 31 which means that x²¹² = x^132 * 31x⁴ = x⁴.

What about the non inversible elements? A weaker version of the previous will still hold and prove the periodicity. Write an element x as (x₁, x₂, x₃, x₄) in . The inversibility case from before is when each x_i ≠ 0. If some of the x_i are 0, then the others will still satisfy that x_i¹³² = 1 by Fermat's little theorem. For example, we have (0, 2, 0, 23)¹³² = (0, 1, 0, 1). So, while x¹³² will not be 1 for these elements, it will still satisfy x^132 * 31x⁴ = x⁴ because the zeroes are multiplying only zeroes and the other values are being multiplied by 1 s.