IMO Team Selection Test 1

Subtracting the second equation from the first yields $x_2-x_3=x_3^2-x_2^2+6x_4(x_3-x_2)$, which we can factor as $0=(x_3-x_2)(x_3+x_2+1+6x_4)$. We see that $x_2=x_3$ or $x_2+x_3+1+6x_4=0$. Similarly, we also have either $x_2=x_3$ or $x_2+x_3+1+6x_1=0$. Hence, if $x_2\ne x_3$, the second equality must hold in both cases; subtracting one from the other, we obtain $x_1=x_4$. We conclude that either $x_2=x_3$ or $x_1=x_4$. Analogously, we get for each permutation $(i,j,k,l)$ of $(1,2,3,4)$ that either $x_i=x_j$ or $x_k=x_l$. We will prove that at least three of the $x_i$ must be equal. If all four are equal, this is true of course. Otherwise, there are two unequal ones, say $x_1\ne x_2$ without loss of generality. Then we have $x_3=x_4$. If also $x_1=x_3$ holds, then there are three equal elements. Otherwise, we have $x_1\ne x_3$, hence $x_2=x_4$ and we also get three equal elements. Up to order, the quadruple $(x_1,x_2,x_3,x_4)$ is thus equal to a quadruple of the shape $(x,x,x,y)$, where we could have that $x=y$. Substituting this in the equations gives $x+y=8x^2$ and $2x=x^2+y^2+6xy$. Adding these two equations: $3x+y=9x^2+y^2+6xy$. The right hand side can be factored as $(3x+y{)}^2$. Defining $s=3x+y$, the equation becomes $s=s^2$, from which we get either $s=0$ or $s=1$. We have $s=3x+y=2x+(x+y)=2x+8x^2$. Hence, $8x^2+2x=0$ or $8x^2+2x=1$. In the first case, we have $x=0$ or $x=-\frac{1}{4}$. We find $y=0-3x=0$ and $y=0-3x=\frac{3}{4}$, respectively. In the second case, we get the factorisation $(4x-1)(2x+1)=0$, hence $x=\frac{1}{4}$ or $x=-\frac{1}{2}$. We find $y=1-3x=\frac{1}{4}$ or $y=1-3x=\frac{5}{2}$, respectively. Altogether, we found the following quadruples: $(0,0,0,0)$, $(-\frac{1}{4},-\frac{1}{4},-\frac{1}{4},\frac{3}{4})$, $(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4})$ and $(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},\frac{5}{2})$, and permutations thereof. It is a simple computation to verify that all these quadruples are indeed solutions to the equations. $\square$