Find the integer values of c 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.
 A: 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?
