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Ukrainian National Mathematical Olympiad

Number theory Difficulty 5.4 AIME, harder Prove it Ukraine

For which positive integers $n \ge 2$ , there exist $n$ odd (not necessarily different) numbers $a_1, a_2, \dots, a_n$ such that
$a_1^2 + a_2^2 + \dots + a_n^2$ is a square of some positive integer?

Solution

Clearly, square of an integer number can give a remainder of $0$ , $1$ or $4$ modulo $8$ . Therefore, only for $n$ of the form $8k + r$ , where $r \in \{0, 1, 4\}$ , such numbers can exist. Let us show how they can be constructed.

$n = 4t,\ a_1 = \dots = a_{n-1} = 1,\ a_n = (2t-1):$
$a_1^2 + a_2^2 + \dots + a_n^2 = (4t-1) \cdot 1^2 + (2t-1)^2 = (2t)^2.$

$n = 8t + 1,\ a_1 = \dots = a_{n-1} = 1,\ a_n = (2t-1):$
$a_1^2 + a_2^2 + \dots + a_n^2 = (8t) \cdot 1^2 + (2t-1)^2 = (2t+1)^2.$

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