**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 a

_{1}, a_{2}, a_{3}, .., a_{n}. By the given property, it follows that:
(a

_{1}– a_{2}a_{3}…a_{n}) = k_{1}n
(a

_{2}– a_{1}a_{3}...a_{n}) = k_{2}n
…

(a

_{n}– a_{1}a_{2}… a_{n-1}) = k_{n}n
where k

_{1}, k_{2}, .., k_{n }are some integers. By multiplying equation 1 above by a_{1}, similarly equation 2 by a_{2}and so on till the last equation above (which would be multiplied by a_{n}), the following set of equations are obtained:
a

_{1}^{2}– C = k_{1}na_{1}
a

_{2}^{2}– C = k_{2}na_{2}
...

a

_{n}^{2}– C = k_{n}na_{n}
where C = a

_{1}a_{2}a_{3}...a_{n}.
Summing the above equations and then rearranging leads to:

(a

_{1}^{2}+ a_{2}^{2}+ … + a_{n}^{2}) = nC + n(k_{1}a_{1}+ k_{2}a_{2}+ … + k_{n}a_{n}) = n(C + k_{1}a_{1}+ k_{2}a_{2}+ … + k_{n}a_{n})
which proves that the sum of squares of the n integers (a

_{1}, a_{2}, a_{3}, .., a_{n}) is a multiple of n.
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