Let's find those values of $c$ for which the equation

$$
x^{5}-5 x+c=0
$$

has two equal roots.
I. solution: Suppose that the number $x=a$ is a double root of the given equation. Then, according to Bézout's theorem, the left side of the equation is divisible by $(x-a)^2$. Already at this point, in the division $\left(x^{5}-5 x+c\right):\left(x^{2}-2 a x+a^{2}\right)$, after the segment of the operation with the constant term of the quotient, the quotient is: $x^{3}+2 a x^{2}+3 a^{2} x+4 a^{3}$ and the remainder is $5\left(a^{4}-1\right) x-\left(4 a^{5}-c\right)$. This polynomial must vanish identically, so $a$ and $c$ must satisfy such relations that both of its coefficients are 0:

$$
5\left(a^{4}-1\right)=0 \quad \text { and } \quad 4 a^{5}-c=0
$$

From the first, $a^{4}=1$, so (restricting to real values) only the numbers

$$
a_{1}=1, \quad a_{2}=-1
$$

can be double roots, and with these, from the second equation, the sought values of $c$ are

$$
c_{1}=4 a_{1}^{5}=4 a_{1}=4, \quad c_{2}=-4
$$

Nagy Elemér (Bp. I., Toldy F. g. IV. o. t.)

II. solution: If $x=a$ is a double root of the given equation, then the left side of the equation can be written as the product of the quadratic $(x-a)^2$ and some cubic polynomial (in which the coefficient of $x^3$ is 1):

$$
\begin{aligned}
& x^{5}-5 x+c=\left(x^{2}-2 a x+a^{2}\right)\left(x^{3}+b x^{2}+d x+e\right): x^{5}+(b-2 a) x^{4}+ \\
& +\left(d-2 a b+a^{2}\right) x^{3}+\left(e-2 a d+a^{2} b\right) x^{2}+\left(-2 a e+a^{2} d\right) x+a^{2} e
\end{aligned}
$$

From the equality of the two sides, the equality of the corresponding coefficients of $a, b, c, d, e$ gives the following five equations:

$$
\begin{aligned}
b-2 a & =0 \\
d-2 a b+a^{2} & =0 \\
e-2 a d+a^{2} b & =0 \\
-2 a e+a^{2} d & =-5 \\
a^{2} e & =c
\end{aligned}
$$

From (1), $b=2 a$, substituting this into (2) gives $d=3 a^{2}$, and with this, from (3) $e=4 a^{3}$, and with $d$ and $e$ from (4) $a^{4}=1$, and finally with $e$ from (5) $c=4 a^{5}$, so by this method we also arrive at the results of the I. solution, and the solution can be concluded as seen there.

Elbert Árpád (Kaposvár, közg. t. IV. o. t.)

III. solution: We will prove that the

$$
g(x)=x^{n}-n a^{n-1} x+a^{n}(n-1)=0
$$

equation has $x=a$ as a double root; in other words: $g(x)$ is divisible by $(x-a)^2$. Transform $g(x)$:

$$
\begin{aligned}
& g(x)=x^{n}-a^{n}-n a^{n-1} x+n a^{n}=\left(x^{n}-a^{n}\right)-n a^{n-1}(x-a)= \\
& =(x-a)\left(x^{n-1}+a x^{n-2}+a^{2} x^{n-3}+\ldots+a^{n-2} x+a^{n-1}-n a^{n-1}\right)
\end{aligned}
$$

The value of the second polynomial in parentheses is 0 when $x=a$, so according to Bézout's theorem, it is divisible by $(x-a)$, hence $g(x)$ is indeed divisible by $(x-a)^2$.

The first two terms on the left side of our given equation are a special case of the first two terms of $g(x)$ for $n=5$ and $-5 a^{4}=-5$, i.e., $a^{4}=1$, or (restricting to real values) $a=1$ and $a=-1$. The theorem's conclusion is valid for our equation if the correspondence also holds for the third terms, i.e., if $c=4 a^{5}$, or $c=4$ and $c=-4$.

Lörinczy László (Szolnok, Verseghy F. g. IV. o. t.) Tatai Péter (Bp. XIV. I. István g. III. o. t.)

Remark: It can generally be proven that the equation $x^{5}+a x+c=0$ has a multiple root if and only if $\left(\frac{c}{4}\right)^{4}+\left(\frac{a}{5}\right)^{5}=0.1$ In our case, $a=-5$, so $c=4$.

Kristóf László (Mosonmagyaróvár, Kossuth L. g. IV. o. t.)[^0]


[^0]:    ${ }^{1}$ See e.g. Szele Tibor: Introduction to Algebra (University Textbook) Tankönyvkiadó, 1953, p. 220.