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define the sequence $\{x_{n}\}$,such $$x_{1}=1,x_{2}=a>1,x_{n+1}=ax_{n}-x_{n-1},n\ge 2,a\in N^{+}$$

show that:this sequence $\{x_{n}\}$ contains an infinite subset such that any two distinct elements of this subset are coprime.

I have see this IMO(1971) problem http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42703,or can see Construction of an infinite set such that any two number from the set are relatively prime But this different problem.

even if I can find closed form $x_{n}$, because $$r^2-ar+1=0,\delta=a^2-4\ge 0$$ because $a\ge 2,a\in N^{+}$,so $$r_{1}=\dfrac{a+\sqrt{a^2-4}}{2},r_{2}=\dfrac{a-\sqrt{a^2-4}}{2}$$ $$\Longrightarrow x_{n}=A(r_{1})^{n-1}+B(r_{2})^{n-1}$$where we can find this two constant $A,B$ by this condition.

but I think this try maybe not usefull to solve this problem

Thank you

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1 Answers1

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I don't think closed form is the best way of going about this.

Rather, say you've picked a few of them already, and assume the lagest one you've picked out is $x_n$. I would like to find an $m > n$ so that $x_m$ is coprime with all earlier picked-out numbers (and thus safe for picking). This is how it can be done:

Note that for any prime $p$, as $i$ varies $x_i \pmod p$ is periodic, starting at $1$ (see below). Let $\{p_1, p_2, \ldots, p_k\}$ be all the primes that divide at least one of the $x_i$ we have picked out. Let $\{T_1, T_2, \ldots, T_k\}$ be the respective minimal periods of $x_i \pmod {p_j}$.

Let $m = \operatorname{lcm}(T_1, T_2, \ldots, T_k) + 1$. Then $x_m$ will be congruent to $1$ modulo any of the primes $p_i$, and thus it is coprime to any of the elements we had earlier chosen.

"Small" addendum on why the sequence $x_i \pmod p$ is periodic, and specifically why it has to return to $1$:

There are only finitely many pairs $(x_n, x_{n-1})$ when viewed modulo $p$, and since those two decide the next number, the sequence has to start repeating itself eventually, but this is not enough to prove that it has to come back to $1$. I have to know that it does, or at least something similar, to guarantee that I avoid several primes with the same period aligning up "badly" (i.e. that eventually at least one of them always divides $x_i$).

If $p\mid a$, then the sequence will be $1, 0, -1, 0, 1, \ldots$, with period $4$, coming back to $1$. Otherwise, note that $p\nmid a$ means the recursion is reversible (given two consecutive elements, you can work your way backwards in the sequence, even $\!\pmod p$), and thus if it eventually repeats, it has to already start in the repeating pattern.

Arthur
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