\(\Delta=m^2-4\ge0\Rightarrow m\ge2\)
Theo Viet: \(\left\{{}\begin{matrix}x_1x_2=1\\x_1+x_2=m\end{matrix}\right.\)
\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=m^2-2\)
\(x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=m^3-3m\)
\(A=x_1^5+x_2^5=\left(x_1^2+x_2^2\right)\left(x_1^3+x_2^3\right)-\left(x_1x_2\right)^2\left(x_1+x_2\right)\)
\(A=\left(m^2-2\right)\left(m^3-3m\right)-m=m\left[\left(m^2-2\right)\left(m^2-3\right)-1\right]\)
\(A=m\left(m^4-5m^2+5\right)\)
- Nếu \(m=5\Rightarrow m^4⋮5\Rightarrow A⋮25\) (thỏa mãn)
- Nếu \(m⋮̸5\Rightarrow\left\{{}\begin{matrix}m⋮̸5\\m^4⋮̸5\Rightarrow m^4-5\left(m^2-1\right)⋮̸5\end{matrix}\right.\)
\(\Rightarrow A⋮̸5\) (ktm)
Vậy \(m_{min}=5\) (là số nhỏ nhất chia hết cho 5 thỏa mãn)
