Find all possible positive integer values of c that $\frac {a^2+b^2+ab} {ab-1}$=c can take in $\mathbb N$. I know that the solutions are c=7 and c=4 but I don't know how to prove this with Vieta Jumping method.

  • 1
    $\begingroup$ What is $a$ and $b$? Are they integers? Or natural numbers? $\endgroup$ – Husain Oct 13 '15 at 15:32
  • $\begingroup$ integers it's a classical problem but I don't know how to solveit by vieta's root jumping $\endgroup$ – Weijie Chen Oct 13 '15 at 16:14
  • $\begingroup$ edited answer to make the Vieta step more prominent. $\endgroup$ – Will Jagy Oct 13 '15 at 19:00


$$ \frac{x^2 + xy + y^2}{xy-1} = c, $$ $$ x^2 + xy + y^2 = c xy - c, $$ $$ x^2 + (1-c)xy + y^2 = -c. $$

We want integers with $$ x^2 + (1-c)xy + y^2 = -c $$ and $c \geq 3,$ because otherwise the quadratic form on the left is positive definite or semidefinite.

There are no solutions with $xy \leq 0$ in any case. we take $x,y > 0.$


We are going to look for solutions that minimize $x + y.$ We can replace $x$ by $$ \color{blue}{x' = (c-1)y - x}. $$ We see that $x' < x$ so $x' + y < x+y,$ unless $2 x \leq (c-1)y. $ We can replace $y$ by $$ \color{blue}{y' = (c-1)x - y}. $$ We see that $y' < y$ so $x + y' < x+y,$ unless $2 y \leq (c-1)x. $


If there are any integer solutions for a particular $c,$ then there are solutions satisfying the inequalities $$ 2 x \leq (c-1)y, $$ $$ 2 y \leq (c-1)x. $$ I'll call these the Hurwitz inequalities and Hurwitz lines. Article in 1907 in German.

For $c=4$ we get $(2,2)$ enter image description here

For (c=7) we get $(2,1), (1,2)$

enter image description here

Let's see: for those who know calculus, the Hurwitz lines intersect this branch of the hyperbola in the minimum value of $x$ and the minimum value of $y.$ What happens when $c \geq 8?$

enter image description here

Well, as you can see in the picture, the branch passes through a point very near $(1,1)$ when $x=y,$ to be specific $$ x = y = \sqrt {\left(1 + \frac{3}{c-3} \right)}, $$ slightly larger then $1.$ However, as soon as $x=2,$ we find the $y$ value along the lower part smaller than $1,$ $$ y = (c-1) - \sqrt {c^2 - 3 c - 3} = \frac{c+4}{ (c-1) + \sqrt {c^2 - 3 c - 3}} $$

This thing has limit $1/2$ as $c \rightarrow \infty.$ $c=8, 0.91723,$ $c=9, 0.85857$ Not a coincidence: $c=7, x=2 \rightarrow y= 1, $ $c=4, x=2 \rightarrow y= 2. $

So that is the proof, for $c \geq 8$ there are no integer lattice points on the arc of the hyperbola within $$ 2 x \leq (c-1)y, $$ $$ 2 y \leq (c-1)x. $$ You can draw graphs for $c = 3,5,6,$ you will see that the arc of the hyperbola between the Hurwitz lines does not contain any lattice points.

I put lots of detail at Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?

  • $\begingroup$ It's very interesting thanks, but I'd like a proof using Vieta's Root Jumping $\endgroup$ – Weijie Chen Oct 13 '15 at 18:40
  • $\begingroup$ @Weijie, it is Veita root jumping. Really really really. Edited to put in the word Vieta in a few places. By the way, root jumping is just one little part of any such argument, everything else is always inequalities, although usually easy ones. $\endgroup$ – Will Jagy Oct 13 '15 at 18:51
  • $\begingroup$ Oh sorry I read it very fast I see it now thanks! $\endgroup$ – Weijie Chen Oct 13 '15 at 19:32
  • $\begingroup$ You said you rarely lie. It means you lie occasionally. xD. By the way, +1. $\endgroup$ – Jaideep Khare Feb 26 '18 at 8:46
  • $\begingroup$ @JaideepKhare I think I recall how this went. Chen, in the first comment above, said he wanted a proof using Vieta Jumping. After that, I put in the first line about not lying. Meanwhile, here is the piece by Hurwitz, which really does help clarify what Vieta Jumping is doing. It is in German, but is about as easy to read as any German mathematics article I have seen. zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf $\endgroup$ – Will Jagy Feb 26 '18 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.