a) Calculate the sum of the squares of the roots of the equation $(x+2)(x-1)(x-3)=2$.

b) Prove that the following equation cannot have all real roots:

$$
x^{10}-2 x^{9}+5 x^{8}+a x^{7}-b x^{6}+c x^{5}+d x^{4}+e x^{3}+f x^{2}+g x+h=0 .
$$
a) By reducing the equation to 0 and breaking it into terms, we can bring it to the form

$$
x^{3}-2 x^{2}-5 x+4=0
$$

If the roots of the equation are $\alpha_{1}, \alpha_{2}, \alpha_{3}$, then the left side must match term by term with the expanded form of

$$
\left(x-\alpha_{1}\right)\left(x-\alpha_{2}\right)\left(x-\alpha_{3}\right)
$$

because the highest degree term of both polynomials is $x^{3}$. Thus,

$$
\begin{gathered}
x^{3}-2 x^{2}-5 x+4-\left(x-\alpha_{1}\right)\left(x-\alpha_{2}\right)\left(x-\alpha_{3}\right)=\left(\alpha_{1}+\alpha_{2}+\alpha_{3}-2\right) x^{2}- \\
-\left(\alpha_{1} \alpha_{2}+\alpha_{1} \alpha_{3}+\alpha_{2} \alpha_{3}+5\right)+\left(\alpha_{1} \alpha_{2} \alpha_{3}+4\right)
\end{gathered}
$$

This difference, on the one hand, takes the value 0 at $\alpha_{1}, \alpha_{2}, \alpha_{3}$, and on the other hand, can be at most a quadratic polynomial, and thus can vanish at most two places if not all the coefficients on the right side are 0. Therefore,

$$
\alpha_{1}+\alpha_{2}+\alpha_{3}=2, \quad \alpha_{1} \alpha_{2}+\alpha_{1} \alpha_{3}+\alpha_{2} \alpha_{3}=-5, \quad \alpha_{1} \alpha_{2} \alpha_{3}=-4
$$

Using the first two, we get

$$
\alpha_{1}^{2}+\alpha_{2}^{2}+\alpha_{3}^{2}=\left(\alpha_{1}+\alpha_{2}+\alpha_{3}\right)^{2}-2\left(\alpha_{1} \alpha_{2}+\alpha_{1} \alpha_{3}+\alpha_{2} \alpha_{3}\right)=2^{2}-2(-5)=14
$$

b) The facts used here are not limited to cubic equations. Generally, if a polynomial of degree $n$

$$
f(x)=a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n-1} x+a_{n}
$$

vanishes at $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$, then it can be rearranged as

$$
a_{0}\left(x-\alpha_{1}\right)\left(x-\alpha_{2}\right) \ldots\left(x-\alpha_{n}\right)
$$

and the following relationships hold between the roots and coefficients:

$$
\begin{aligned}
a_{1} & =-a_{0}\left(\alpha_{1}+\alpha_{2}+\ldots+\alpha_{n}\right) \\
a_{2} & =a_{0}\left(\alpha_{1} \alpha_{2}+\alpha_{1} \alpha_{3}+\ldots \alpha_{n-1} \alpha_{n}\right) \\
a_{3} & =-a_{0}\left(\alpha_{1} \alpha_{2} \alpha_{3}+\alpha_{1} \alpha_{2} \alpha_{4}+\ldots+\alpha_{n-2} \alpha_{n-1} \alpha_{n}\right), \ldots \\
a_{n} & =(-1)^{n} a_{0} \alpha_{1} \alpha_{2} \ldots \alpha_{n}
\end{aligned}
$$

Generally, $a_{k}$ can be obtained by selecting $k$ of the $\alpha$'s in every possible way, multiplying them, and then multiplying the sum of the products by $a_{0}$ or $(-a_{0})$ depending on whether $k$ is even or odd.

If, specifically, the roots of equation (1) are $\beta_{1}, \beta_{2}, \ldots, \beta_{10}$, to calculate their sum of squares, it is sufficient to know the coefficients of the 9th and 8th degree terms:

$$
\begin{gathered}
\beta_{1}+\beta_{2}+\ldots+\beta_{10}=-2 \\
\beta_{1} \beta_{2}+\ldots+\beta_{1} \beta_{10}+\beta_{2} \beta_{3}+\ldots+\beta_{9} \beta_{10}=5
\end{gathered}
$$

Thus,

$$
\begin{aligned}
\beta_{1}^{2} \beta_{2}^{2}+\ldots+\beta_{10}^{2} & =\left(\beta_{1}+\beta_{2}+\ldots+\beta_{10}\right)^{2}-2\left(\beta_{1} \beta_{2}+\beta_{1} \beta_{3}+\ldots+\right. \\
& \left.+\beta_{9} \beta_{10}\right)=(-2)^{2}-2 \cdot 5=-6
\end{aligned}
$$

This is impossible in the realm of real numbers, so it cannot be that equation (1) has 10 real roots, which is what we needed to prove. Horányi Sándor (Budapest, Móricz Zs. g. IV. o. t.)

Note. We assumed that our equations have as many roots as their degree. In the case of (a), this can be seen from the fact that the left side of the reduced equation takes the values $-2, 4, -6, 16$ at $-2, 0, 2, 4$ respectively. These values have alternating signs, so between each pair of adjacent points, it takes the value 0.

Part (b) can be considered an indirect proof: If there were 10 (real) roots, their sum of squares would have to be positive, which is impossible. If we rely on the knowledge of complex numbers as well, it is known that every (at least first-degree) polynomial has a root in the complex numbers. From this, it follows - using [3], that if a polynomial vanishes at $\alpha$, the factor $x-\alpha$ can be factored out - that a polynomial of degree $n$ has $n$ roots in the complex numbers, counting multiplicities. Thus, equation (1) also has 10 roots in the complex numbers, but they cannot all be real because their sum of squares is negative. (It is possible that none of them are real.)[^0]

[^0]:    ${ }^{1}$ See, for example, János Surányi: Polynomial Identity, K. M. L. 23 (1961/11.), pp. 103-105.

    ${ }^{2}$ We arrive at these by introducing a new number $i$, the imaginary unit, whose square is -1. Its real multiples are the imaginary numbers, and the sum of a real and an imaginary number gives the complex numbers. In the realm of complex numbers, the basic operations can be defined such that the usual operational rules remain valid for them. In the realm of complex numbers, every (at least first-degree) polynomial has a root. See, for example, Tamás Varga: Comprehensive Mathematics for Maturity, Tankönyvkiadó, Budapest, 1961, 11th edition, pp. 55-57; and Richárd Rieger: Complex Numbers, Középiskolai Szakköri Füzetek, Tankönyvkiadó, Budapest, 1954.

    ${ }^{3}$ See, for example, the article cited in the first footnote.