Let $n\in\Bbb{N}$ such that $n = a^2+b^2+c^2$ for some natural numbers $a, b, c$. Prove that $$9n = (p_1a + q_1b + r_1c)^2 + (p_2a + q_2b + r_2c)^ 2 + (p_3a + q_3b + r_3c)^ 2$$ where $p_j, q_j, r_j$ are all nonzero integers.
The question asks me to prove $9a^2+9b^2+9c^2 = (p_1a + q_1b + r_1c)^2 + (p_2a + q_2b + r_2c)^ 2 + (p_3a + q_3b + r_3c)^ 2$.
If for some $(p_j, q_j, r_j)\in \Bbb{Z}$, $$3a=p_1a + q_1b + r_1c$$$$3b=p_2a + q_2b + r_2c$$ $$3c=p_3a + q_3b + r_3c$$ Then the given equation will automatically be proven correct.
The first equation gives $(3a-p_1)a = q_1b + r_1c$. This diophantine equation solution iff $(b, c) | (3a-p_1)a$. This condition will be easily satisfied for many $p_1 , q_1, r_1$. Same goes for second and third equations.
This is my proof.
Is this proof ? The question seems too easy to be a Olympiad question. I think it has some mistakes. Can anyone please verify it ?
This question is taken from INMO 2007.