det(A² + B²) is positive for commuting real matrices

Idea. Over $\mathbb{R}$, $A^2+B^2$ does not factor, but over $\mathbb{C}$ it does, provided $A$ and $B$ commute. The two factors are complex conjugates, so their determinants are conjugate and multiply to a modulus squared.

Step 1: the factorisation. Regard $A,B$ as matrices in $M_n(\mathbb{C})$. Then
$$(A+iB)(A-iB)=A^2-iAB+iBA+B^2=A^2+B^2,$$
because $-iAB+iBA=-i(AB-BA)=0$. This is where the commutation is used.

Step 2: determinants. Multiplicativity of $\det$ over $\mathbb{C}$ gives
$$\det(A^2+B^2)=\det(A+iB)\,\det(A-iB).$$

Step 3: conjugates. Since $A$ and $B$ have real entries, the matrix $A-iB$ is the entrywise complex conjugate of $A+iB$. The determinant is a polynomial with real (integer) coefficients in the entries, so $\det(\bar M)=\overline{\det M}$. Hence
$$\det(A^2+B^2)=\det(A+iB)\overline{\det(A+iB)}=|\det(A+iB)|^2\ge0.$$

Remark. Without commutation the statement is false. For example, for $n=2$ take $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $B=\begin{pmatrix}0&0\\1&0\end{pmatrix}$: then $A^2=B^2=0$ and the determinant is $0$, which does not contradict; but with $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, $B=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ one gets $A^2+B^2=2I$, $\det=4\ge0$ again. Counterexamples exist for $n\ge2$ (e.g. suitable non-commuting nilpotent-plus-diagonal pairs), but the point of the problem is only that the proof breaks at Step 1.

Lesson. Complexify to factor sums of squares; real matrices produce conjugate factors, hence a non-negative product.