\[chris\] Let ${\Sigma}$ be a topological sphere and $(I_i,II_i)$, $i=1,2$, two Codazzi pairs with mean curvature $H_i$ and extrinsic curvature $K_i$. If both pairs have the same third fundamental form with $K_i(p)\neq 0$ for all $p\in{\Sigma}$ and $\frac{H_1}{K_1}=\frac{H_2}{K_2}$, then $(I_1,II_1)=(I_2,II_2)$.

When we particularize this result to the case of two isometric immersions from a Riemannian sphere in $\r^3$ satisfying the assumptions above, we get an easy proof of the uniqueness to the Christoffel problem. Besides this proof is original in the sense that the classical approaches to this problem use integration theory on surfaces (see [@H; @Sp]).

It is known [@Mi] that if $(I_i,II_i)$ is a Codazzi pair with non vanishing extrinsic curvature, then $(III_i,II_i)$ is also a Codazzi pair with mean curvature $\frac{H_i}{K_i}$. Consequently, from Corollary \[bonnet\] we deduce that $II_1=II_2$.

Thus, since $K(III_i,II_i)=\frac{1}{K_i}$ we have that $K_1=K_2$. Finally, using that $I_i=-\frac{1}{K_i}\,III_i+2\frac{H_i}{K_i}\,II_i$, it follows that $I_1=I_2$.

We observe that the previous proof is based in the simple fact that $(III,II)$ is a Codazzi pair. That is, the Bonnet theorem in $\r^3$ and the theorem of uniqueness of the Christoffel problem are a direct consequence of the Abstract Bonnet Theorem, when it is applied to the Codazzi pair $(I,II)$ or 