Let $F = \mathbb{Q}\setminus\{-1,0,1\}$ and $f$ the function defined over F by $f(x) = \dfrac{x^2-1}{x}$. Show that : $$ \bigcap_{n\ge 1} f^n(F)= \emptyset$$

I don't have many ideas for this problem. I tryed solving $f(x) = a$ over F and the existence of a solution is equivalent to $a^2 + 4$ being a perfect square. But it doesn't really get me anywhere. Does anyone have an idea ?


I believe the following works. At a high level, we argue that if there are terms in the intersection then there must be a term with smallest denominator, and that term turns out to imply that a $\frac 1 c$ term must be in the image with minimal $c.$ But this is impossible.

There are no nontrivial integers a such that $a^2 + 4$ is a perfect square, so we know that there are no integers in the infinite intersection. Let $a = \frac b c$ be in the intersection, with $c>1$ minimal and $b,c$ relatively prime. Since $a$ is in the intersection, it has a preimage, so $$\frac b c = f(\frac d e) = \frac d e - \frac e d = \frac{d^2 - e^2}{de},$$ with $d,e$ relatively prime as well. We can also ask that $\frac d e$ be in the intersection, since for example $a$ must be in $f^{(n+1)}$ for every $n,$ so $e \geq c > 1.$

Now, I claim that $gcd(d^2 - e^2, de) = 1,$ so that the right hand side above is a reduced fraction, and $de = c.$ This is because $$gcd(d^2-e^2, de) = gcd(d^2-e^2, d) gcd(d^2-e^2,e) = gcd(e^2, d) gcd(d^2,e) = 1$$ (we use that $gcd(x, \cdot)$ is multiplicative here).

So we get that $\frac b c = \frac{d^2 - e^2} {de}$ are both reduced, in particular $c= de.$ Since we picked $c$ minimal, $e \geq c,$ and therefore $c = e$ and $d = 1.$ So we have that $b = 1 - c^2,$ or $a = \frac 1 c - \frac c 1 = f(-c) = f(\frac 1 c)$ for an integer $c.$ (note, these are the only $2$ preimages as $f$ is essentially quadratic: if $f(x) = a$ then $x$ must satisfy the quadratic formula for $x^2 - ax -1,$ as suggested in the quesiton.)

We go one step further back. $-c$ isn't in the image of $f,$ so that can't be how we arrived at $a$ in the image of $f^{(3)}$ (say). Therefore $\frac 1 c$ must be in the image of $f.$ We repeat the argument above, to find a $\frac g h$ such that $f(\frac g h) = \frac 1 c = \frac{g^2-h^2}{gh},$ again this is reduced and so we have $g = 1, h = c$ from the denominator, but then $1 - c^2 = 1$, which is impossible.

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    $\begingroup$ Very nice solution thanks ! If anyone is struggling to understand why we can get $e\ge c$ because c is minimal, notice that we can always pick $d/e$ such that it is in the intersection. $\endgroup$ – aleph0 Jun 15 at 12:20
  • $\begingroup$ Ah, yes. Thank you for pointing this out. I've clarified a line accordingly. This was a fun problem. $\endgroup$ – Artimis Fowl Jun 15 at 12:33

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