Find the integer values of c

Question

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.

Answer 1

THIS IS VIETA JUMPING. I RARELY LIE ABOUT MATHEMATICS.

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

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 $x,y > 0.$

THIS IS THE PART WHERE VIETA ROOT JUMPING IS EXPLICIT!!!!!!!

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. $

SEE. VIETA ROOT JUMPING. RIGHT THERE ABOVE.

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 $(2,2)$

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

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 $c \geq 8?$

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?