6.245. Find the coefficients $m$ and $n$ of the quadratic trinomial $x^{2}+m x+n$, given that the remainders when it is divided by the binomials $x-m$ and $x-n$ are respectively $m$ and $\boldsymbol{n}$.
Solution.

Dividing $x^{2}+m x+n$ by $x-m$, we have the solution

$$
\begin{aligned}
& \begin{array}{c||c}
x^{2}+m x+n & x-m \\
-x^{2}-m x & \frac{x+2 m}{}
\end{array} \\
& \begin{array}{l}
2 m x+n \\
\frac{2 m x+2 m^{2}}{2 m^{2}}+n
\end{array}
\end{aligned}
$$

Thus, $2 m^{2}+n=m$.

Dividing $x^{2}+m x+n$ by $x-n$, we get

$$
\begin{aligned}
& -\frac{x^{2}+m x+n}{-\frac{x^{2}-n x}{(m+n) x}} \quad \frac{x-n}{x+(m+n)} \\
& -\frac{(m+n) x-n(m+n)}{n+n(m+n)}
\end{aligned}
$$

From this, we have $n+n(m+n)=n$.

We obtain the system of equations

$$
\left\{\begin{array} { l } 
{ 2 m ^ { 2 } + n = m , } \\
{ n + n ( m + n ) = n }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2 m^{2}-m+n=0, \\
n(m+n)=0
\end{array} \Rightarrow n=0 \text { or } n+m=0\right.\right.
$$

Therefore, this system is equivalent to two systems of equations:

$$
\left\{\begin{array} { l } 
{ 2 m ^ { 2 } - m + n = 0 , } \\
{ n = 0 }
\end{array} , \text { or } \left\{\begin{array} { l } 
{ 2 m ^ { 2 } - m + n = 0 , } \\
{ m + n = 0 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ m _ { 1 } = 0 , } \\
{ n _ { 1 } = 0 ; }
\end{array} \left\{\begin{array} { l } 
{ m _ { 2 } = \frac { 1 } { 2 } , } \\
{ n _ { 2 } = 0 ; }
\end{array} \left\{\begin{array}{l}
m_{3}=1, \\
n_{3}=-1
\end{array}\right.\right.\right.\right.\right.
$$

Answer: $(0 ; 0),\left(\frac{1}{2} ; 0\right),(1 ;-1)$.