If the sum of two sectional curvatures is invariant then M is conformally flat

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My problem is:

At any $x \in M$, the condition $$K(X_1,X_2)+K(X_3,X_4)=K(X_1,X_3)+K(X_2,X_4)$$ for $X_1,X_2,X_3,X_4 \in T_{x}M$, where the vectors are pairwise orthogonal and $K$ is the sectional curvature of $M$. Then $M^{n}$ is conformally flat, where $n \ge 4$.

Definition: $L(X,Y)=\frac{1}{n-2}(\mbox{Ric}(X,Y)-\frac{1}{2}ns\langle X, Y \rangle)$ is the Schouten tensor of $M$, where $s$ is the scalar curvature of $M$.

Definiton: $$\langle C(X,Y)Z,W \rangle =\langle R(X,Y)Z,W \rangle-L(X,W)\langle Y,Z\rangle -L(Y,Z)\langle X,W \rangle +L(X,Z)\langle Y,W \rangle +L(Y,W)\langle X,Z \rangle$$ is the Weyl tensor of $M$.

To show the problem it suffices to show that $C=0$. So, \begin{align*} \langle C(X_1,X_2)X_2,X_1 \rangle &=K(X_1,X_2)-L(X_1,X_1)-L(X_2,X_2)\\ \langle C(X_3,X_4)X_4, X_3 \rangle &=K(X_3,X_4)-L(X_3,X_3)-L(X_4,X_4)\\ \langle C(X_1,X_3)X_3,X_1 \rangle&=K(X_1,X_3)-L(X_1,X_1)-L(X_3,X_3)\\ \langle C(X_2,X_4)X_4,X_2 \rangle &= K(X_2, X_4)-L(X_2,X_2)-L(X_4,X_4), \end{align*} implies \begin{equation*} \langle C(X_1,X_2)X_2, X_1 \rangle +\langle C(X_3,X_4)X_4,X_3 \rangle = \langle C(X_1,X_3)X_3,X_1 \rangle +\langle C(X_2,X_4)X_4,X_2 \rangle. \end{equation*} If I chosen $X_1=X_3=X$ and $X_2=X_4=Y$ we have $$\langle C(X,Y)Y,X \rangle=0.$$ However from that point I can't get $C=0.$ How can I do this?

Thank you for your answer.