A simple number theory problem

Divisibility puzzle: Suppose that there are n integers which have the following
property: the difference between the product of any n - 1 integers and
the remaining one is divisible by n . Prove that the sum of the square of
these n numbers is also divisible by n.

Solution:

Denote the n integers by a₁, a₂, a₃, .., a_n. By the given property, it follows that:

(a₁ – a₂ a₃…a_n) = k₁n

(a₂ – a₁ a₃ ...a_n) = k₂n

…

(a_n – a₁ a₂… a_n-1) = k_nn

where k₁, k₂, .., k_n are some integers. By multiplying equation 1 above by a₁, similarly equation 2 by a₂ and so on till the last equation above (which would be multiplied by a_n), the following set of equations are obtained:

a₁² – C = k₁na₁

a₂² – C = k₂na₂

...

a_n² – C = k_nna_n

where C = a₁ a₂ a₃ ...a_n.

Summing the above equations and then rearranging leads to:

(a₁²+ a₂² + … + a_n²) = nC  + n(k₁a₁ + k₂a₂ + … + k_na_n)  = n(C + k₁a₁ + k₂a₂ + … + k_na_n)

which proves that the sum of squares of the n integers (a₁, a₂, a₃, .., a_n) is a multiple of n.