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.



Strong divisibility of Lucas sequences
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](http://mathworld.wolfram.com/PrimitivePrimeFactor.html) 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 $u\_m = u\_d$.


