Task B-2.7. Determine the value of the parameter $m \in \mathbb{Z}$ for which the quadratic equation

$$
(m-1) x^{2}+(2 m-1) x+3=0
$$

has rational solutions.
## First Solution.

For the solutions of a quadratic equation to be rational, its discriminant must be the square of some integer. Let this be $k^{2}$.

$$
\begin{aligned}
& D=(2 m-1)^{2}-4(m-1) \cdot 3=4 m^{2}-16 m+13 \\
& 4 m^{2}-16 m+13=k^{2}
\end{aligned}
$$

Let's solve the obtained quadratic equation for $m$.

$$
\begin{aligned}
4 m^{2} & -16 m+13-k^{2}=0 \\
m_{1,2} & =\frac{16 \pm \sqrt{256-16\left(13-k^{2}\right)}}{8}= \\
& =\frac{16 \pm \sqrt{48+16 k^{2}}}{8}= \\
& =\frac{16 \pm 4 \sqrt{3+k^{2}}}{8}= \\
& =\frac{4 \pm \sqrt{3+k^{2}}}{2}
\end{aligned}
$$

The solution is an integer only if $3+k^{2}$ is the square of some integer, i.e.,

$$
\begin{aligned}
& 3+k^{2}=n^{2} \\
& n^{2}-k^{2}=3 \\
& (n-k)(n+k)=1 \cdot 3
\end{aligned}
$$

This is possible only for

$$
\left\{\begin{array} { l } 
{ n - k = \pm 1 } \\
{ n + k = \pm 3 }
\end{array} \quad \text { or } \left\{\begin{array}{l}
n-k= \pm 3 \\
n+k= \pm 1
\end{array}\right.\right.
$$

From these systems and under the condition $n^{2}>k^{2}$, it follows that $k^{2}=1$.

Then $m=1$ or $m=3$.

For $m=1$, the equation is not quadratic, so the only solution is $m=3$.