Library / Indian National Mathematical Olympiad /74 of 103

Indian National Mathematical Olympiad

Number theory Difficulty 6.5 National Olympiad Prove it India

Problem:
Let $n$ be a natural number such that $n = a^{2} + b^{2} + c^{2}$ , for some natural numbers $a, b, c$ . Prove that
$9 n = (p_{1} a + q_{1} b + r_{1} c)^{2} + (p_{2} a + q_{2} b + r_{2} c)^{2} + (p_{3} a + q_{3} b + r_{3} c)^{2}$
where $p_{j}$ 's, $q_{j}$ 's, $r_{j}$ 's are all nonzero integers. Further, if $3$ does not divide at least one of $a, b, c$ , prove that $9 n$ can be expressed in the form $x^{2} + y^{2} + z^{2}$ , where $x, y, z$ are natural numbers none of which is divisible by $3$ .

Solution

Solution:
It can be easily seen that
$9 n = (2 b + 2 c - a)^{2} + (2 c + 2 a - b)^{2} + (2 a + 2 b - c)^{2}$
Thus we can take $p_{1} = p_{2} = p_{3} = 2$ , $q_{1} = q_{2} = q_{3} = 2$ and $r_{1} = r_{2} = r_{3} = -1$ . Suppose $3$ does not divide $\gcd(a, b, c)$ . Then $3$ does divide at least one of $a, b, c$ ; say $3$ does not divide $a$ . Note that each of $2 b + 2 c - a$ , $2 c + 2 a - b$ and $2 a + 2 b - c$ is either divisible by $3$ or none of them is divisible by $3$ , as the difference of any two sums is always divisible by $3$ . If $3$ does not divide $2 b + 2 c - a$ , then we have the required representation. If $3$ divides $2 b + 2 c - a$ , then $3$ does not divide $2 b + 2 c + a$ . On the other hand, we also note that
$9 n = (2 b + 2 c + a)^{2} + (2 c - 2 a - b)^{2} + (-2 a + 2 b - c)^{2} = x^{2} + y^{2} + z^{2}$
where $x = 2 b + 2 c + a$ , $y = 2 c - 2 a - b$ and $z = -2 a + 2 b - c$ . Since $x - y = 3(b + a)$ and $3$ does not divide $x$ , it follows that $3$ does not divide $y$ as well. Similarly, we conclude that $3$ does not divide $z$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.