Chinese Mathematical Olympiad

Suppose that $r$ is any positive integer. Since there are infinitely many primes, there is a prime $p$, such that $$p>(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2).①$$ Because $p$ is prime and ①, we have $(p,a_1^2+a_2^2+a_3^2)=1$. $p$ is coprime to $p-1$; from the Chinese remainder theorem, there is a positive integer $n$ such that $$n\equiv r(\mathrm{mod}p-1),$$ $$n(a_1^r+a_2^r+a_3^r)+a_1^r-a_3^r\equiv 0(\mathrm{mod}p).$$ From the above and Fermat's theorem, $$(n+1)a_1^r+n{a}_2^r+(n-1)a_3^r\equiv n(a_1^r+a_2^r+a_3^r)+a_1^r-a_3^r\equiv 0(\mathrm{mod}p).$$ From the assumption of the problem, $$(n+1)b_1^r+n{b}_2^r+(n-1)b_3^r\equiv 0(\mathrm{mod}p).$$ Again from above and Fermat's little theorem, $$n(b_1^r+b_2^r+b_3^r)+b_1^r-b_3^r\equiv 0(\mathrm{mod}p).$$ Eliminate $n$ from the two congruences: $$(a_1^r+a_2^r+a_3^r)(b_1^r-b_3^r)\equiv (b_1^r+b_2^r+b_3^r)(a_1^r-a_3^r)(\mathrm{mod}p).$$ From the choice of $p$, this congruence is actually an equality: $$(a_1^r+a_2^r+a_3^r)(b_1^r-b_3^r)=(b_1^r+b_2^r+b_3^r)(a_1^r-a_3^r).$$ Thus, $$(a_2{b}_1{)}^r+2(a_3{b}_1{)}^r+(a_3{b}_2{)}^r=(a_1{b}_2{)}^r+2(a_1{b}_3{)}^r+(a_2{b}_3{)}^r.$$ We then prove the following lemma.

Lemma Assume that $x_1,\dots ,x_s,y_1,\dots ,y_s$ are real numbers, $$0<x_1\le x_2\le \cdots \le x_s,0<y_1\le y_2\le \cdots \le y_s,$$ such that for any positive integers $r$, $$x_1^r+x_2^r+\cdots +x_s^r=y_1^r+y_2^r+\cdots +y_s^r.$$ Then $x_i=y_i$ for $i=1,2,\dots ,s$.

Proof of the lemma. We use induction on $s$. If $s=1$, take $r=1$; then $x_1=y_1$. Assume that the lemma holds when $s=t$. When $s=t+1$, if $x_{t+1}\ne y_{t+1}$, say $x_{t+1}<y_{t+1}$, $${\left(\frac{x_1}{y_{t+1}}\right)}^r+\cdots +{\left(\frac{x_{t+1}}{y_{t+1}}\right)}^r={\left(\frac{y_1}{y_{t+1}}\right)}^r+\cdots +{\left(\frac{y_t}{y_{t+1}}\right)}^r+1\ge 1.$$ Because $0<\frac{x_i}{y_{t+1}}<1$ ($1\le i\le t+1$), take the limit $r\to +\infty$, and we have $0\ge 1$, a contradiction. So $x_{t+1}=y_{t+1}$, and then $x_1^r+\cdots +x_t^r=y_1^r+\cdots +y_t^r$, $r=1,2,\dots$. By induction the lemma holds for all positive integer $s$.

Now return to the proof of the problem. Since $a_1,a_2,a_3,b_1,b_2,b_3$ are distinct, $$a_2{b}_1\ne a_3{b}_1,a_3{b}_1\ne a_3{b}_2,a_1{b}_2\ne a_1{b}_3,$$ $$a_1{b}_3\ne a_2{b}_3,a_2{b}_1\ne a_2{b}_3.$$ From the previous equality and the lemma we know that $$a_2{b}_1=a_1{b}_2,a_3{b}_1=a_1{b}_3,a_3{b}_2=a_2{b}_3.$$ Then $\frac{b_1}{a_1}=\frac{b_2}{a_2}=\frac{b_3}{a_3}$. Write $\frac{b_1}{a_1}=\frac{k}{l}$, $(k,l)=1,l\ge 1$; then $b_i=\frac{k}{l}{a}_i$, $i=1,2,3$. From $2b_1+b_2=\frac{k}{l}(2a_1+a_2)$ and the assumption of the problem (with $n=1$), $2a_1+a_2\mid 2b_1+b_2$ and $\frac{k}{l}$ is an integer. So $l=1$ and $b_i=k{a}_i$, $i=1,2,3$.