= Solution
Take a code alphabet of size $D\geq2$ and prescribed positive <integer> lengths $\ell_1,\ldots,\ell_q$. A <decipherable code> means that concatenation of codewords determines the entire finite sequence of codewords uniquely. First prove the necessary <Kraft inequality>. If $A(m,L)$ counts sequences of $m$ codewords with total length $L$, <unique decodability> gives $A(m,L)\leq D^L$. Therefore
$$
K^m:=\left(\sum_{i=1}^qD^{-\ell_i}\right)^m
=\sum_{L=m\ell_{\min}}^{m\ell_{\max}}A(m,L)D^{-L}
\leq m(\ell_{\max}-\ell_{\min})+1.
$$
Taking $m$th roots and letting $m\to\infty$ gives $K\leq1$.
Conversely, order the lengths increasingly and construct a <prefix code> in the rooted $D$-ary <tree>. At depth $\ell_j$, previously selected codewords forbid exactly $\sum_{i<j}D^{\ell_j-\ell_i}$ vertices. The partial <Kraft inequality> implies
$$
\sum_{i<j}D^{\ell_j-\ell_i}\leq D^{\ell_j}-1,
$$
so an available vertex exists. Select it as the next codeword. Earlier lengths are no greater, so this preserves the <prefix code> property. A <prefix code> has <unique decodability>: in two purported parsings, the first unequal codewords would force one to be a prefix of the other. This proves the <equivalence of decipherable and prefix code lengths>: \b[the two existence conditions are equivalent].
For a one-symbol alphabet, at most one positive-length codeword can be decipherable: two different unary codewords commute under concatenation. At most one is also exactly the prefix-code condition, so the equivalence still holds. The same argument covers countably many prescribed positive lengths for $D\geq2$: necessity applies to each finite subset, and $K\leq1$ ensures finitely many words at each length, allowing the ordered construction. Empty codewords are excluded, as required for <unique decodability> of arbitrary sequences.
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