# Iterations of function over rationals

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.
• 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. – aleph0 Jun 15 at 12:20