TITLE: Strong divisibility of Lucas sequences
QUESTION [6 upvotes]: Let $a$ and $b$ be relatively prime integers and let $u_n$ be their associate Lucas sequence, i.e., the second order linear recurrence sequence satisfying $u_0 = 0$, $u_1 = 1$ and $u_{n+2} = au_{n+1} + bu_n$, for each nonnegative integer $n$.
It is well know that $(u_n)_{n=0}^\infty$ is a strong divisibility sequence, i.e., it holds 
$$(\bullet) \quad \gcd(u_m, u_n) = u_{\gcd(m,n)} ,$$
for all the integers $m,n \geq 0$ (put $\gcd(0,0) := 0$). 
This in turn implies that
$$(\star) \quad m \mid n \Rightarrow u_m \mid u_n ,$$
for all the integers $m,n \geq 0$.
My question is: Are there some nice hypothesis under which also the reverse implication holds in ($\star$) ?
Note that from ($\bullet$), we get
$$ u_m \mid u_n \Rightarrow u_m = \gcd(u_m, u_n) = u_{\gcd(m,n)} ,$$
so if $(u_n)_{n=0}^\infty$ is injective then $m = \gcd(m,n)$ and thus $m \mid n$. So my question can be also answered if one gets some nice hypothesis under which $(u_n)_{n=0}^\infty$ is injective.
Thank you in advance for any suggestion.

REPLY [4 votes]: I found a method to solve this problem. We recall the primitive prime factor theorem
Theorem 2.3.1 (Florian Luca, Effective methods for diophantine equations)
If $k \notin \{1,2,3,4,6\}$, then $u_k$ has a primitive prime factor except when $(a,\Delta,k)$, where $\Delta = a^2 + 4b$, is one of the following triples:
$$(1, 5, 5), (1, -7, 5), (2, -40, 5), (1, -11,5), (1, -15, 5), (12, -76, 5), (12, -1364, 5),$$
$$(1, -7, 7), (1, -19, 7),$$
$$(2, -24, 8), (1, -7, 8),$$
$$(2,-8,10), (5, -3, 10),$$
$$(1, 5, 12), (1, -7, 12), (1, -11, 12), (2, -56, 12), (1, -15, 12), (1, -19, 12),$$
$$(1, -7, 13),$$
$$(1, -7, 18),$$
$$(1, -7, 30).$$
Now, let $m$ and $n$ be positive integers such that $u_m \mid u_n$ but $m \nmid n$.
From $(\bullet)$ we have that
$$u_m = \gcd(u_m, u_n) = u_d ,$$
where $d = \gcd(m,n) < m$. It follows that $u_m$ has not a primitive prime factor.
From Theorem 2.3.1 then or $m \in \{1,2,3,4,6\}$ or $(a,\Delta,m)$ is one of the triples listed above.
In each of these cases, we can (patiently) check if actually there exist or not a divisor $d$ of $m$ such that $u_m = u_d$.