South African Mathematics Olympiad Third Round

The quadratic formula gives us $$\frac{-a\pm \sqrt{a^2-4a+4b}}{2}$$ and $$\frac{-a\pm \sqrt{a^2-4a-4b}}{2}$$

Suppose that the four consecutive numbers are $n-1,n,n+1,n+2$. The parabola $y=x^2+ax+a$ reaches its minimum at $x=-a/2$, and the line $x=-a/2$ is its axis of symmetry. The two pairs of solutions both have to have this axis of symmetry, and the solutions to the second equation have to lie closer to the minimum at $x=-a/2$. Thus the only possibility is that $n-1$ and $n+2$ are the solutions to the first equation, while $n$ and $n+1$ are the solutions to the second equation. This gives us the equations $$(n-1{)}^2+a(n-1)+a=b,(1)$$ $$n^2+an+a=-b,(2)$$ $$(n+1{)}^2+a(n+1)+a=-b.(3)$$ Subtract (1) from (3) to obtain $$4n+2a=-2b.(4)$$ Subtract (2) from (3) to obtain $$2n+1+a=0.(5)$$ Multiply (5) by 2 and subtract from (4): $$-2=-2b,$$ so $b=1$. Now we know that the two roots of $x^2+ax+a=1$ differ by 3: $$\frac{-a+\sqrt{a^2-4a+4}}{2}=\frac{-a-\sqrt{a^2-4a+4}}{2}+3,$$ so $$\sqrt{a^2-4a+4}=\sqrt{(a-2{)}^2}=3,$$ which means that $a-2=\pm 3$. Once again, we find that the two possibilities are $a=-1,b=1$ and $a=5,b=1$.