For which value of the parameter $k$ can the expression

$$
(3 k-2) x(x+k)+k^{2}(k-1)
$$

be written in the form $(a x+b)^{2}$?
The task is to find such values of $k$ for which there exist suitable $a$ and $b$ such that the following identity holds for all values of $x$:

$$
(3 k-2) x(x+k)+k^{2}(k-1) \equiv(a x+b)^{2}
$$

or, reduced to 0 and arranged according to the powers of $x$:

$$
\left(3 k-2-a^{2}\right) x^{2}+[k(3 k-2)-2 a b] x+k^{2}(k-1)-b^{2} \equiv 0
$$

This can only hold for all values of $x$ if the coefficients of $x^2$ and $x$, as well as the constant term, are each 0, because otherwise the left side would be a quadratic or linear polynomial, or a non-zero constant, which can be 0 for at most two values of $x$. Thus, we need to find a triplet of real numbers $k, a, b$ such that

$$
\begin{gathered}
3 k-2=a^{2} \\
k(3 k-2)=2 a b \\
k^{2}(k-1)=b^{2}
\end{gathered}
$$

We can eliminate $a$ and $b$ simultaneously by subtracting the square of (3) from four times the product of (2) and (4):

$$
\begin{gathered}
4(3 k-2) k^{2}(k-1)-k^{2}(3 k-2)^{2}=0 \\
k^{2}(3 k-2)(k-2)=0
\end{gathered}
$$

This only holds for $k_{1}=0, k_{2}=2 / 3$, and $k_{3}=2$. However, the left side of (2) is negative for $k_{1}$, so no real value can be obtained for $a$; for $k_{2}$, $a=0$, and the given expression becomes $-4 / 27$, which cannot be written as the square of a real expression, but it does not contain $x$ and thus would not be considered a solution to the problem. For $k_{3}$, from (2) we get $a= \pm 2$, and from (3) we get $b=4 / a= \pm 2$. Since the expression $(a x+b)^{2}$ is a square, the two signs of $a$ do not yield essentially different expressions; we can therefore require that $a$ be positive, for example; thus, from $a=2$ we get $b=2$.

Accordingly, for $k=2$, the given expression can be written as $(2 x+2)^{2}=4(x+1)^{2}$.

Kelemen György (Tamási, Béri Balogh Á. g. IV. o. t.)

Remarks. 1. We do not arrive at another solution by seeking which condition ensures that the zeros of the left and right sides of (1) coincide, i.e., the roots of the equation obtained by reducing to 0, and by writing the equality of their sum and product expressible by the coefficients. In fact, this line of thought - although it leads to (5) - is incomplete without supplementation, as it only ensures the equality of the ratio of the coefficients, not the identity of the expressions.

2. The same deficiency applies to the following line of thought: The equation $(a x+b)^{2}=0$ has two equal roots, and this must also be the case for the equation obtained by reducing the left side of (1) to 0, so its discriminant must be equal to 0.