1961 IMO Problems/Problem 1

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Problem

(Hungary) Solve the system of equations:

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

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

Solution

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

When $a \neq 0$ , we divide $(**)$ by $(*)$ 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 $a$ and $a^2 > b^2$ . Now, we have $x+y = \frac{a^2 + b^2}{2a}$ ; $xy = \left(\frac{a^2 - b^2}{2a}\right)^2$ , so $x,y$ 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 $(3a^2 - b^2), (3b^2 - a^2)$ were simultaneously negative, then their sum, $2(a^2 + b^2)$ , would also be negative, which cannot be. Therefore our quadratic's discriminant is positive when $3a^2 > b^2$ and $3b^2 > a^2$ . But we have already replaced the first inequality with the sharper bound $a^2 > b^2$ . 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 $a$ is positive and $3b^2 > a^2 > b^2$ . Q.E.D.