South African Mathematics Olympiad

Solution 1: If we subtract the two equations, we obtain $$a{x}^2+8-8x^2-a=(a-8)(x^2-1)=(a-8)(x+1)(x-1)=0,$$ thus either $a=8$ or $x=-1$ or $x=1$. If $a=8$, we are left with $$5x^3+8x^2+8=(x+2)(5x^2-2x+4)=0.$$ The second factor has no real roots, since its discriminant $2^2-4\cdot 5\cdot 4=-76$ is negative. Thus $x=-2$ in this case. If $x=-1$, we get $a=-5x^3-8x^2=-3$, and if $x=1$, we get $a=-5x^3-8x^2=-13$. In summary, there are three possible pairs: $(a,x)=(8,-2)$, $(a,x)=(-3,-1)$ and $(a,x)=(-13,1)$.

Solution 2: Solving the second equation for $a$ gives us $a=-5x^3-8x^2$, thus $$5x^3+(-5x^3-8x^2)x^2+8=-5x^5-8x^4+5x^3+8=0.$$ The polynomial can be factorised: $$\begin{array}{rl}-5x^5-8x^4+5x^3+8& =5x^3(1-x^2)+8(1-x^4)=5x^3(1-x^2)+8(1+x^2)(1-x^2)\\ & =(1-x^2)(5x^3+8x^2+8)=(1-x)(1+x)(x+2)(5x^2-2x+4).\end{array}$$ Again, we find that $x=1$, $x=-1$ or $x=-2$, since the quadratic factor has no real roots. The value of $a$ is obtained from the equation $a=-5x^3-8x^2$, so we end up with the three possibilities $(a,x)=(8,-2)$, $(a,x)=(-3,-1)$ and $(a,x)=(-13,1)$ again.