Iranian Mathematical Olympiad

Lemma 1. Let $A\equiv 2(\mathrm{mod}3)$ be a positive integer. Then there exists a prime number $p$ such that $p\equiv 2(\mathrm{mod}3)$ and $p^{\alpha }\mid A$ where $\alpha$ is an odd integer.

Proof of lemma. Assume to the contrary that there is not such a prime number $p$. Therefore, if $A=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_k^{{\alpha }_k}$ is the prime factorization of $A$, we have two cases for $p_i$'s, $1\le i\le k$, to consider:

Case 1. $p_i\equiv 1(\mathrm{mod}3)\Rightarrow p_i^{{\alpha }_i}\equiv 1(\mathrm{mod}3)$.

Case 2. $p_i\equiv 2(\mathrm{mod}3)$ and $2|{\alpha }_i\Rightarrow p_i^{{\alpha }_i}\equiv (p_i^2{)}^{\frac{{\alpha }_i}{2}}\equiv 1(\mathrm{mod}3)$.

As a result, we must have $A\equiv 1(\mathrm{mod}3)$ which is a contradiction. $\square$

Lemma 2. Let $p\equiv 2(\mathrm{mod}3)$ be a prime number. Show that $\{0^3,1^3,\dots ,(p-1{)}^3\}$ is a complete residue system modulo $p$.

Proof of lemma. Obviously $i^3\equiv 0^3(\mathrm{mod}p)$ iff $i\equiv 0(\mathrm{mod}p)$. Suppose that $p\nmid i,j$. Our goal is to show that $i^3\equiv j^3(\mathrm{mod}p)$ iff $i\equiv j(\mathrm{mod}p)$. One part of the proof is obvious. To prove the other part, suppose that $p=3t+2$. Then by Fermat's Little Theorem, we have $i^{3t+1}\equiv j^{3t+1}\equiv 1(\mathrm{mod}p)$. Hence, we have $$i^{3t}i\equiv i^{3t+1}\equiv j^{3t+1}\equiv (j^3{)}^{t}j\equiv i^{3t}j(\mathrm{mod}p).$$ Since $(i,p)=1$, we get $i\equiv j(\mathrm{mod}p)$. $\square$

Now we are ready to solve the main problem. We claim that there is no such triple. Assume to the contrary that $$a^2+b^2+c^2=2013k(ab+bc+ca)$$ for some positive integer $k$.

First, without loss of generality we can suppose that $a$, $b$ and $c$ have no common factor, because if $(a,b,c)=d>1$, we can divide them by $d$ to get a new triple with no common factor. We have $(a+b+c{)}^2=(2013k+2)(ab+bc+ca)$. $2013k+2\equiv 2(\mathrm{mod}3)$, so by lemma 1 there is some prime number $p\equiv 2(\mathrm{mod}3)$ such that $p^{2n+1}\nmid 2013k+2$ ($n\ge 0$). $$\begin{array}{rl}{p}^{2n+1}\nmid 2013k+2& \Rightarrow p^{2n+1}|(a+b+c{)}^2\Rightarrow p^{2n+2}|(a+b+c{)}^2\\ & \Rightarrow p^{2n+2}|(2013k+2)(ab+bc+ca)\Rightarrow p|ab+bc+ca.\end{array}$$ As a result, $p|a+b+c$ and $p|ab+bc+ca$. Hence $$\begin{array}{rl}0& \equiv ab+bc+ca\equiv ab+c(a+b)\equiv ab+c(-c)(\mathrm{mod}p)\Rightarrow ab\equiv c^2(\mathrm{mod}p)\\ & \Rightarrow c^3\equiv abc(\mathrm{mod}p).\end{array}$$ By a similar argument, $a^3\equiv b^3\equiv abc(\mathrm{mod}p)$, so by lemma 2 we deduce $a\equiv b\equiv c(\mathrm{mod}p)$ and since $p|a+b+c$ and $3\nmid p$, we find that $p$ divides $a$, $b$ and $c$, which contradicts our assumption that $(a,b,c)=1$. $\square$