Mean squared error for an estimator is $\sigma^4\{a^2(n^2-1)-2a(n-1)+1\},$ where $n$ is the sample size and $a$ is constant. I have to show this mean squared error(mse) is minimized at $a=\frac{1}{(n+1)}.$ So, I set mse equal to $0$, that is,
$$\sigma^4\{a^2(n^2-1)-2a(n-1)+1\}=0$$ $$\Rightarrow a^2(n^2-1)-2a(n-1)+1=0$$ $$\Rightarrow a=\frac{-\{-2(n-1)\}\pm \sqrt{\{-2(n-1)\}^2-4(n^2-1)(1)}}{2(n^2-1)}$$ $$\Rightarrow a=\frac{2(n-1)\pm \sqrt{4(n-1)^2-4(n-1)(n+1)}}{2(n^2-1)}$$ $$\Rightarrow a=\frac{2(n-1)\pm \sqrt{4(n-1)\{(n-1)-(n+1)\}}}{2(n^2-1)}$$ $$\Rightarrow a=\frac{2(n-1)\pm \sqrt{4(n-1)(-2)}}{2(n^2-1)}$$
But I don't end up with the result $a=\frac{1}{(n+1)}.$ Could you please tell me where I am doing mistake?