RMO 2015

We start by checking trivial solutions when $a=0, \pm 1$ , $b=0, \pm 1$ , and $c=0,\pm1$ . This is not so hard, but it takes a while, and I'll omit the process here. There are eight distinct solutions, which are $(a,b,c)= (0,1,-1),(0,-1,1), (-1,0,1),(-1,1,0),(-1,-1,0), (1,1,0),(1,-1,0),(1,0,-1).$ Now, we will prove that there is no more solution. Assume that $a,b,c \not= 0,\pm 1$ . From both equation, we have that $abc=a^3-a=b^3-b.$ The latter equation can be written as $(a-b)(a^2+ab+b^2-1)=0,$ so, there are two cases. First, if $a=b$ , it implies from the original equation that $a(c-a)=-1$ , and it results in $a=\pm 1$ , which has already considered. Therefore, we assume that $a \not= b$ , and what we have left is the case when $a^2+ab+b^2=1,$ or $(a+b)^2=ab+1$ . The difference between those two original equations is $(b-a)c = (a-b)(a+b).$ Since $a \not= b$ , we have that $a+b=-c$ . Put this into $(a+b)^2=ab+1$ , and get $c^2=ab+1$ . We now have $a+b+c=0$ , and also $\begin{aligned} a^2 &= bc+1,\\ b^2 &= ca+1,\\ c^2 &= ab+1. \end{aligned}$ Sum them up, and we get $a^2+b^2+c^2=ab+bc+ca+3$ . With the fact that $a+b+c=0$ , we have $a^2+b^2+c^2=-2(ab+bc+ca)$ . Therefore, $a^2+b^2+c^2=2.$ However, this equation has no solution other than what we have derived previously, because if at least one of $|a|,|b|,|c| \ge 2$ , the right hand side of the equation will exceed $2$ .