Infinite sequence that must be constant Let $c$ be a positive integer and $a_1,a_2, \ldots, a_n$ an infinite sequence of positive integers, such that $a_{n+1}=\sqrt{a_n^3-ca_n}$. Prove that $(a_n)$ is eventually constant, that is there exists a positive integer $N$ such that $a_N=a_{N+i}$ for all $i \in \mathbb{N}$.
-This was given to me by a friend. Up to now, I have no solution. What I have shown is that actually what we are to prove is equivalent to $(a_n)$ being constant, and if we suppose that $(a_n)$ is not constant, I have proved that it has to be stricly increasing.
 A: Observe that $$a_2>a_1\iff a_1>\frac{1+\sqrt{4c+1}}{2}$$
Now we have $a_2=\sqrt{a_1(a_1^2-c)}>a_1>\frac{1+\sqrt{4c+1}}{2}$, because $a_1^2-c>a_1$. Thus we also have $a_3>a_2$. Continuing this way we have that the sequence is strictly increasing. 
Similarly we can show that $a_2<a_1$ implies the sequence is strictly decreasing. But a sequence of positive integers can not be strictly decreasing. 
If we have $a_1=a_2$ we are already done.
So the sequence is strictly increasing. Now observe that if $p\mid a_n, p\nmid c$ for some prime $p$, we will have that $v_p(a_{n+1})=\frac{v_p(a_n)}{2}$. But this can not continue forever. Hence we must have $p\mid a_n\implies p\mid c$. This readily implies that the set of prime divisors of the set $\{a_1,a_2,\ldots\}$ is finite. But we also have that the sequence is strictly increasing. Hence $p^{\alpha}\mid a_n, p^{\alpha}\nmid c$ for some $n\in \mathbb{N}$. Now let us look at $v_p(a_{n+1})$. We can easily see that $v_p(a_n)>v_p(a_{n+1})=\frac{v_p(a_n)+v_p(c)}{2}>v_p(c)$. This implies $v_p(a_m)$ will not be integer for some large enough $m$, a contradiction. A contradiction.
Hence we must have $a_1=a_2$. Thus we are done.
