Let $A$ and $B$ be real symmetric matrices. Suppose that for all integers $k\ge1$ we have $$ {\rm tr}((A+B)^k)={\rm tr}(A^k) + {\rm tr}(B^k). $$ Prove that then, $AB=0$ (the converse is clearly true as $A$ and $B$ are symmetric). Note: this question is NOT a duplicate of a resembling question asked on this forum. We do NOT suppose that
$$ {\rm tr}((xA+yB)^k)=x^k{\rm tr}(A^k) +y^k{\rm tr}(B^k). $$ for all $x,y$ real, but ONLY for $x=y=1$.