3. Determine all pairs $(a, b)$ of real numbers for which the quadratic trinomials $P(x)=$ $=x^{2}+a x+b$ and $Q(x)=x^{2}+b x+a$ have the following property: each of the equations

$$
a P(x)+b Q(x)=0 \quad \text{and} \quad a Q(x)+b P(x)=0
$$

is a quadratic equation with a double root.

(Jaroslav Švrček)
SOLUTION. Substituting the trinomials $P$ and $Q$ into the given equations

$a\left(x^{2}+a x+b\right)+b\left(x^{2}+b x+a\right)=0, \quad$ resp. $\quad a\left(x^{2}+b x+a\right)+b\left(x^{2}+a x+b\right)=0$

we rewrite them in standard form

$$
(a+b) x^{2}+\left(a^{2}+b^{2}\right) x+2 a b=0, \quad \text { resp. } \quad(a+b) x^{2}+2 a b x+\left(a^{2}+b^{2}\right)=0 .
$$

We see that in the case $a+b=0$ these equations are not quadratic, which contradicts the problem statement. The sought numbers $a, b$ must therefore satisfy the condition $a+b \neq 0$, which we will assume is satisfied from now on.

As we well know, quadratic equations have double roots precisely when their discriminants are zero. In the case of equations (1), by writing out their discriminants, we obtain for the unknowns $a, b$ the system of equations

$$
\begin{aligned}
& \left(a^{2}+b^{2}\right)^{2}-4(a+b) \cdot 2 a b=0, \\
& (2 a b)^{2}-4(a+b)\left(a^{2}+b^{2}\right)=0,
\end{aligned}
$$

which we rewrite in the form

$$
\begin{aligned}
\left(a^{2}+b^{2}\right)^{2} & =8(a+b) a b \\
(a+b)\left(a^{2}+b^{2}\right) & =a^{2} b^{2}
\end{aligned}
$$

Multiply the first equation by the expression $a^{2}+b^{2}$ (non-zero due to the condition $a+b \neq 0$). We get the equation

$$
\left(a^{2}+b^{2}\right)^{3}=8\left(a^{2}+b^{2}\right)(a+b) a b .
$$

We can substitute $\left(a^{2}+b^{2}\right)(a+b)$ with $a^{2} b^{2}$ in the right-hand side according to the second equation from (2). After this simplification, the previous equation takes the form

$$
\left(a^{2}+b^{2}\right)^{3}=8 a^{3} b^{3} \quad \text { or } \quad\left(a^{2}+b^{2}\right)^{3}=(2 a b)^{3} \text {. }
$$

Since the third powers of two real numbers are known to be equal precisely when their bases are equal, from the derived equation (3) it follows that $a^{2}+b^{2}=2 a b$ or $(a-b)^{2}=0$, which occurs only in the case $a=b$. Substituting into the first equation from (2), we get

$$
\left(a^{2}+a^{2}\right)^{2}=8(a+a) a^{2} \text {, after simplification } a^{3}(a-4)=0 .
$$

Thus, only two possibilities arise - the pair $(a, b)$ is either $(0,0)$ or $(4,4)$. Since the pair $(0,0)$ contradicts the derived condition $a+b \neq 0$, only the pair $(a, b)=(4,4)$ is feasible. This pair is indeed a solution to our problem, as it is a solution to the system (2), which expresses the zero discriminants of both equations.*[^1]

NOTE. The key step in the given solution was deriving the consequence (3) from the system of equations (2) - see also the introductory problem N2. Otherwise, the consequence (3) can also be obtained by multiplying the equations from (2) and then dividing the resulting equality by the expression $a+b$, which we already know is not equal to zero.