1961 IMO Problems/Problem 1

Problem

(Hungary) Solve the system of equations:

$\begin{matrix} \quad x + y + z \!\!\! &= a \; \, \\ x^2 +y^2+z^2 \!\!\! &=b^2 \\ \qquad \qquad xy \!\!\! &= z^2 \end{matrix}$

where $\displaystyle a$ and $\displaystyle b$ are constants. Give the conditions that $\displaystyle a$ and $\displaystyle b$ must satisfy so that $\displaystyle x, y, z$ (the solutions of the system) are distinct positive numbers.

Solution

Note that $\displaystyle x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2$ , so the first two equations become

$\begin{matrix} \quad (x + y) + z \!\!\! &= a \; \; (*) \\ (x+y)^2 - z^2 \!\!\! &=b^2 (**) \end{matrix}$ .

We note that $\displaystyle (x+y)^2 - z^2 = \Big[ (x+y)+z \Big]\Big[ (x+y)-z\Big]$ , so if $\displaystyle a$ equals 0, then $\displaystyle b$ must also equal 0. We then have $\displaystyle x+y = -z$ ; $\displaystyle xy = (x+y)^2$ . This gives us $\displaystyle x^2 + xy + y^2 = 0$ . Mutiplying both sides by $\displaystyle (x-y)$ , we have $\displaystyle x^3 - y^3 = 0$ . Since we want $\displaystyle x,y$ to be real, this implies $\displaystyle x = y$ . But $\displaystyle x^2 + x^2 + x^2$ can only equal 0 when $\displaystyle x=0$ (which, in this case, implies $\displaystyle y,z = 0$ ). Hence there are no positive solutions when $\displaystyle a = 0$ .

When $\displaystyle a \neq 0$ , we divide $\displaystyle (**)$ by $\displaystyle (*)$ to obtain the system of equations

$\begin{matrix} (x+y)+z &= a \; \quad \\ (x+y)-z &= b^2/a \end{matrix}$ ,

which clearly has solution $x+y = \frac{a^2 + b^2}{2a}$ , $z = \frac{a^2 - b^2}{2a}$ . In order for these both to be positive, we must have positive $\displaystyle a$ and $\displaystyle a^2 > b^2$ . Now, we have $x+y = \frac{a^2 + b^2}{2a}$ ; $xy = \left(\frac{a^2 - b^2}{2a}\right)^2$ , so $\displaystyle x,y$ are the roots of the quadratic $m^2 - \frac{a^2 + b^2}{2a}m + \left(\frac{a^2 - b^2}{2a}\right)^2$ . The discriminant for this equation is

$\left(\frac{a^2 + b^2}{2a}\right)^2 - \left(2\frac{a^2 -b^2}{2a}\right)^2 = \frac{ (3a^2 - b^2)(3b^2 - a^2) }{4a^2}$ .

If the expressions $\displaystyle (3a^2 - b^2), (3b^2 - a^2)$ were simultaneously negative, then their sum, $\displaystyle 2(a^2 + b^2)$ , would also be negative, which cannot be. Therefore our quadratic's discriminant is positive when $\displaystyle 3a^2 > b^2$ and $\displaystyle 3b^2 > a^2$ . But we have already replaced the first inequality with the sharper bound $\displaystyle a^2 > b^2$ . It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from $\left(\frac{a^2 + b^2}{2a}\right)^2 > \left(\frac{a^2 + b^2}{2a}\right)^2 - \left(2\frac{a^2 -b^2}{2a}\right)^2$ or from Descartes' Rule of Signs). We have now found the solutions to the system, and determined that it has positive solutions if and only if $\displaystyle a$ is positive and $\displaystyle 3b^2 > a^2 > b^2$ . Q.E.D.