1997 JBMO Problems/Problem 5

Problem

Let $n_1$ , $n_2$ , $\ldots$ , $n_{1998}$ be positive integers such that $n_1^2 + n_2^2 + \cdots + n_{1997}^2 = n_{1998}^2.$ Show that at least two of the numbers are even.

Solution

In order to prove that at least two numbers are even, we need to prove that it is impossible to have exactly one number even and that it is impossible to have exactly no numbers even.

Lemma 1: Impossible to have exactly 1 number even
Assume one of the numbers is even. If $n_{1998}$ is even, then the left side is congruent to $1$ modulo $2$ while the right side is congruent to $0$ modulo $2,$ which can not happen. If one of $n_1, n_2, \cdots n_{1997}$ is even, then the left side is congruent to $0$ modulo $2$ while the right side is congruent to $1$ modulo $2,$ which can not happen. Thus, it is impossible for exactly one of the numbers to be even.

Lemma 2: Impossible to have all numbers odd
Assume all numbers are odd (or that there are no even numbers). If $n_i$ is odd for $1 \le i \le 1998,$ then $n_i^2 \equiv 1 \pmod{8}.$ That means the left side is congruent to $5$ modulo $8$ while the right side is congruent to $1$ modulo $8.$ Since this can not happen, it is impossible for none of the numbers to be even.

Since it’s impossible to have exactly one or zero even numbers, at least two of the integers must be even.

1997 JBMO (Problems • Resources)
Preceded by Problem 4	Followed by Last Problem
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1997 JBMO Problems/Problem 5

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