3. For what value of the parameter m is the sum of the squares of the roots of the equation

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

the smallest?
3. $\mathrm{m}=3$.

Note that the discriminant $\mathrm{D}$ of the equation is equal to

$$
\mathrm{D}=(\mathrm{m}+1)^{2}-4(\mathrm{~m}-1)=(\mathrm{m}-1)^{2}+4>0
$$

therefore, the equation has two roots for any $\mathrm{m}$. Let's represent the sum of the squares of the roots $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ as

$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}
$$

From Vieta's theorem, we get

$$
x_{1}^{2}+x_{2}^{2}=\frac{(m+1)^{2}}{4}-2(m-1)
$$

The minimum value of $x_{1}^{2}+x_{2}^{2}$ is achieved at the minimum point of the function $\mathrm{f}(\mathrm{m})=(\mathrm{m}+1)^{2}-8(\mathrm{~m}-1)$, which is $\mathrm{m}_{\min }=3$.