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Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

For which values of parameter aa the system of equations
{x+y+z=0(xy+yz)+axz=0 has a unique solution? \begin{cases} x + y + z = 0 \\ (xy + yz) + a xz = 0 \end{cases} \text{ has a unique solution?}

Solution

It is evident that for every value of parameter aa there exists solution x=y=z=0x = y = z = 0, and so all we need is to find out when this solution is unique.

From the first equation we have that y=xzy = -x - z. Substitute (xz)(-x - z) instead of yy into the second equation:

x22xzz2+axz=0-x^2 - 2xz - z^2 + a xz = 0

or

x2+xz(2a)+z2=0x^2 + xz(2 - a) + z^2 = 0.

We have obtained a quadratic equation with respect to xx. Its discriminant D=z2(2a)24z2=z2(4aa2)D = z^2(2 - a)^2 - 4z^2 = z^2(4a - a^2).

If z=0z = 0 then the system has zero solution, thus for all z0z \neq 0 our quadratic equation shouldn't have any solutions, that is D=z2(4aa2)<0D = z^2(4a - a^2) < 0.

And the last inequality holds for a(0,4)a \in (0,4).

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