I'm supposed to find the value of $m$ that minimises the MSE$(\hat{\sigma}^{2}_{m})$ that is defined as
$$\frac{2(n-k)\sigma^{4}}{m^{2}}+\bigg(\frac{(n-k)\sigma^{2}}{m}- \sigma^{2}\bigg )^{2}$$
I rewrote this:
$$\frac{2(n-k)\sigma^{4}}{m^{2}}+ \frac{(n-k-m)^{2}\sigma^{4}}{m^{2}} = \frac{\sigma^{4}(2(n-k)+(n-k-m)^{2})}{m^{2}}$$
Then differentiating this with respect to $m$ gave me: $$\frac{-2\sigma^{4}(2(n-k)+(n-k-m)^{2})}{m^{3}} = \frac{2\sigma^{4}(2(k-n)-(n-k-m)^{2})}{m^{3}}$$
However the answer was: $$\frac{2\sigma^{4}(k-n)(k-n+m-2)}{m^{3}}$$
I tried moving stuff around to get to the answer but it did not work, so I think I differentiated it wrongly.
If you take derivative directly: $$\left[\frac{2(n-k)\sigma^{4}}{m^{2}}+\bigg(\frac{(n-k)\sigma^{2}}{m}- \sigma^{2}\bigg )^{2}\right]'=\\ -\frac{4(n-k)\sigma ^4}{m^3}+2\left(\frac{(n-k)\sigma^{2}}{m}- \sigma^{2}\right)\cdot \left(-\frac{(n-k)\sigma ^2}{m^2}\right)=\\ \frac{4(k-n)\sigma ^4}{m^3}+\frac{2(k-n)(n-k-m)\sigma ^4}{m^3}=\\ \frac{2(k-n)[2+n-k-m]\sigma ^4}{m^3},$$ which is sign-different from your answer.