Let $M$ be a regular surface without umbilical points. Prove $M$ is minimal iff its Gauss map $\mathrm N$ satisfies for all $p\in M$ and $w_1,w_2\in \mathrm T_pM$ the equality $$ \left\langle \mathrm d_p\mathrm N w_1,\mathrm d_p\mathrm Nw_2 \right\rangle_{\mathrm Np} = \lambda (p) \left\langle w_1,w_2\right\rangle_p $$ for some $\lambda(p)\neq 0$ depending on $p$ only.
My attmept. By self adjointness of the shape operator $$ \left\langle \mathrm d_p\mathrm N w_1,\mathrm d_p\mathrm Nw_2 \right\rangle_{\mathrm Np} = \left\langle \mathrm (\mathrm d_p\mathrm N)^2 w_1,w_2 \right\rangle_p.$$ Now, $$ \left\langle \mathrm (\mathrm d_p\mathrm N)^2 w_1,w_2 \right\rangle_p=\lambda (p) \left\langle w_1,w_2\right\rangle_p$$ for all $w_1,w_2$ iff $(\mathrm d_p\mathrm N)^2=\lambda (p)I$ iff $k_1^2=k_2^2=\lambda(p)$.
Under the assumption $M$ has no umbilical points, the principal curvatures are distinct, so this is equivalent to $k_1=-k_2$.
Is my solution correct? I'm looking at another student's solution and see it's a long computation in terms of bases of tangent spaces and hope it's not necessary.