Problem
Given three spin-1/2 particles with the total spin operator $\vec{S}=\sum\limits_{i=1}^3 \vec{S}_i$ and its $z$ projection $S_z=\sum\limits_{i=1}^3 S_{z,i}$, and the Hamiltonian
$$H = J\sum\limits_{i=1}^3 \vec{S}_i \cdot \vec{S}_{i+1} $$ (assuming for $i=3$ that $i+1=1$), calculate the eigenstates and the eigenenergies.
Hint: Rewrite $H$ as a function of $S^2$ and $S_i^2$.
Work
I've already calculated the basis for $\vec{S}^2$ and $S_z$
$$ \vert 3/2,3/2\rangle \equiv \vert\uparrow\uparrow\uparrow \rangle \\ \vert3/2,1/2\rangle \equiv \frac{1}{\sqrt{3}}\Big( \vert\downarrow\uparrow\uparrow \rangle + \vert\uparrow\uparrow\downarrow \rangle + \vert\uparrow\downarrow\uparrow \rangle \Big)\\ \vert3/2,-1/2\rangle \equiv \sqrt{\frac{2}{3}}\Big( \vert\downarrow\downarrow\uparrow \rangle + \vert\downarrow\uparrow\downarrow \rangle + \vert\uparrow\downarrow\downarrow \rangle \Big)\\ |3/2,-3/2\rangle \equiv \vert\downarrow\downarrow\downarrow \rangle \\ $$ with eigenvalues according to
$$S^2\vert s,m \rangle = \hbar^2s(s+1)\vert s,m \rangle = \frac{15\hbar^2}{4}\vert 3/2,m \rangle \\ S_z\vert s,m \rangle = \hbar m\vert 3/2,m \rangle \text{ for } m=-\frac{3}{2},-\frac{1}{2},\frac{1}{2}\frac{3}{2}. $$
I'm now attempting to rewrite the Hamiltonian according to the hint, using $$\vec{S}_i \cdot \vec{S}_{i+1} = \frac{1}{2}\Big[\Big(\vec{S}_i + \vec{S}_{i+1}\Big)^2 - \Big(\vec{S}_i^2 + \vec{S}_{i+1}^2\Big) \Big].$$
Issue
I'm not certain I'm applying the hint correctly. With the above, $$H = \frac{J}{2}\Big[ \Big(\vec{S}_1 + \vec{S}_2 \Big)^2 + \Big(\vec{S}_1 + \vec{S}_3 \Big)^2 + \Big(\vec{S}_2 + \vec{S}_3 \Big)^2 -2\Big( \vec{S}_1^2 + \vec{S}_2^2 +\vec{S}_3^2\Big) \Big],$$
which once expanded and using the expansion of $\vec{S}^2 = \Big(\vec{S}_1 + \vec{S}_2 +\vec{S}_3\Big)^2$ gets me to
$$H = J\sum\limits_{i=1}^2 S_i^2,$$
which just seems wrong to me, since it's not written as a function of $S^2$ and $S_i^2$, as the hint suggests.