Vieta jumping with non-monic polynomials

I have recently discovered Vieta jumping as a problem-solving technique. In order to teach myself about it, I have located most (all of?) the standard references, both here on MSE and "out there" (via web searches).

However, all the examples I've found are of a form where the equation to be subjected to root-flipping is a monic [quadratic] polynomial, e.g., $$x^2+rx+s = 0.$$ This, of course, is quite useful, because then Vieta's root equation $$x_2 = -\frac{B}{A} - x_1 = -r-x_1$$ implies that $x_2$ must be an integer when $r$ and $x_1$ are integers. On the other hand, when $A > 1$ we have instead $$x_2 = -\frac{B}{A} - x_1 = -\frac{r}{t}-x_1$$ for some integer $t > 1$, so one can't immediately conclude that $x_2$ is an integer. (I am currently trying to solve a problem just like this.)

Does anyone have an example of a problem solved by Vieta jumping [at least partly], where the root-flipping equation is not monic?

Thanks, Kieren.

• – Will Jagy May 17 '14 at 18:44
• math.stackexchange.com/questions/632097/… – Will Jagy May 17 '14 at 18:49
• Hi Will! Thanks for the comments… but I don't see where the non-monic polynomials are. In the Markov example, we have $3xyz \mid (x^2+y^2+z^2)$, so in any variable, the root-flipping equation is monic, e.g., $$x^2 - 3xyz + (y^2+z^2)=0.$$ Same goes for the quartic. Or am I missing something? – Kieren MacMillan May 17 '14 at 18:50
• Here's the concrete example I'm working with: math.stackexchange.com/questions/538131/… It was in one of the comments (by Jack D'Aurizio) that I first heard about Vieta jumping. Using it, I can get the root-flipping equation $$kx^2 + 2b(k-3b)x + (kb^2-2b^3+1)=0.$$ I can obtain the desired result $(a,b)=(4,1)$ if I assume $x_2$ is an integer. But I'm stuck on the rational case. – Kieren MacMillan May 17 '14 at 18:56