= Solution
Let each mass be $m$ and each spring constant be $k$, with the $x_i$ denoting longitudinal displacements from equilibrium. The <Lagrangian> is
$$
L=\frac m2\sum_{i=1}^4\dot x_i^2
-\frac k2\left[(x_2-x_1)^2+(x_3-x_2)^2+(x_4-x_3)^2\right].
$$
The inverse coordinate transformation is
$$
x_1=q_1+q_3,\quad x_4=q_1-q_3,\quad
x_2=q_2+q_4,\quad x_3=q_2-q_4.
$$
Substitution gives
$$
\boxed{
L=m(\dot q_1^2+\dot q_2^2+\dot q_3^2+\dot q_4^2)
-k\left[(q_2-q_1)^2+(q_4-q_3)^2+2q_4^2\right]}.
$$
Thus the <Euler-Lagrange equations> separate into a symmetric pair,
$$
m\ddot q_1+k(q_1-q_2)=0,
\qquad
m\ddot q_2+k(q_2-q_1)=0,
$$
and an antisymmetric pair,
$$
m\ddot q_3+k(q_3-q_4)=0,
\qquad
m\ddot q_4+k(3q_4-q_3)=0.
$$
The symmetric <normal modes> are:
* $q_1=q_2$, corresponding to $(x_1,x_2,x_3,x_4)\propto(1,1,1,1)$ and the translational frequency
$$
\boxed{\omega_0=0};
$$
* $q_1=-q_2$, corresponding to $(1,-1,-1,1)$ and
$$
\boxed{\omega_{\rm s}=\sqrt{\frac{2k}{m}}}.
$$
For antisymmetric modes, the squared dimensionless frequencies are the eigenvalues of
$$
\begin{pmatrix}1&-1\\-1&3\end{pmatrix},
$$
namely $2\mp\sqrt2$. They give
$$
\boxed{\omega_-=\sqrt{\frac{k}{m}(2-\sqrt2)}},
\qquad
(q_3,q_4)\propto(1,\sqrt2-1),
$$
and
$$
\boxed{\omega_+=\sqrt{\frac{k}{m}(2+\sqrt2)}},
\qquad
(q_3,q_4)\propto(1,-1-\sqrt2).
$$
In the original coordinates their mode shapes are respectively
$$
(1,\sqrt2-1,1-\sqrt2,-1)
$$
and
$$
(1,-1-\sqrt2,1+\sqrt2,-1).
$$
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