= Solution
For a lattice of period $a$, wavevectors differing by the reciprocal-lattice vector $2\pi/a$ describe the same translation eigenvalue. The first <Brillouin zone> may therefore be chosen as
$$
-\frac{\pi}{a}\leq k<\frac{\pi}{a}.
$$
<Bloch theorem> states that every energy eigenstate in a <periodic potential> may be chosen in the form
$$
\boxed{\psi_k(x)=e^{ikx}u_k(x),\qquad u_k(x+a)=u_k(x)}.
$$
To prove it, let $T_a$ be translation by one period:
$$
(T_a\psi)(x)=\psi(x+a).
$$
Periodicity gives $[T_a,H]=0$. The translation operator is unitary, so it can be diagonalized within each energy eigenspace. Choose a common eigenstate,
$$
T_a\psi=\lambda\psi.
$$
Unitarity gives $|\lambda|=1$, so $\lambda=e^{ika}$ for some real $k$. Thus $\psi(x+a)=e^{ika}\psi(x)$. Defining $u_k(x)=e^{-ikx}\psi(x)$ gives
$$
u_k(x+a)=e^{-ik(x+a)}e^{ika}\psi(x)=u_k(x),
$$
which proves the theorem.
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