4. Different real numbers $a, b, c$ are such that among the three equations

$$
x^{2}+a x+b=0, \quad x^{2}+b x+c=0, \quad x^{2}+c x+a=0
$$

any two have exactly one common root. Find $a^{2}+b^{2}+c^{2}$.

## Senior League
4. Note that the common root of the first two equations is also the root of their difference. We get that it is equal to $p=\frac{c-b}{a-b}$; similarly for the other common roots: $q=\frac{a-c}{b-c}$ and $r=\frac{b-a}{c-a}$. Among numbers of this form, there must be two of different signs, that is, $p, q, r$ are not all the same.

Note that if $p=q$, then $p=q=r$, but this is impossible. Therefore, $p, q, r$ are pairwise distinct, and the quadratic polynomials can be represented in the form: $(x+p)(x+r),(x+q)(x+p),(x+r)(x+q)$; by Vieta's theorem

$$
\left\{\begin{array}{l}
a=p+q=q r \\
b=q+r=r p \\
c=r+p=p q
\end{array}\right.
$$

We get $2(p+q+r)=p q+q r+r p ; p-r=r(q-p)$. The last equation can be multiplied by two similar ones, and we obtain $p q r=1$. Also from the system, we get the equation $p=q(r-1)$. Multiplying it by similar ones, we get $(p-1)(q-1)(r-1)=1$. Expanding the brackets, it is easy to obtain $p+q+r=-1$ and $p q+q r+r p=-2$. Further

$$
\begin{aligned}
a^{2}+b^{2}+c^{2} & =2\left(p^{2}+q^{2}+r^{2}\right)+2(p q+q r+r p)= \\
& =2(p+q+r)^{2}-2(p q+q r+r p)=6
\end{aligned}
$$

## Senior League