Using one integer point to find other integer points on a hyperbola

Question

Is it possible to use an integer point on a hyperbola of the form $x^2-y^2 = a$ to find other integer points on the hyperbola? For example if we have the parabola $x^2-y^2 = 221$ and we know that $(\pm 111, \pm 110)$ are on the curve, can we use this information to find the remaining integer points?

Answer 1

Notice that $221=13 \times 17$ and $x^2-y^2=(x-y)(x+y)$, so we don't have many possibilities:

$x-y=1$ implies $x+y=221$ and here we get the solution $(111,110)$

$x-y=-1$ implies $x+y=-221$ and here we get the solution $(-111,-110)$

$x-y=13$ implies $x+y=17$ and here we get the solution $(15,2)$

$x-y=-13$ implies $x+y=-17$ and here we get the solution $(-15,-2)$

$x-y=17$ implies $x+y=13$ and here we get the solution $(15,-2)$

$x-y=-17$ implies $x+y=-13$ and here we get the solution $(-15,2)$

$x-y=221$ implies $x+y=1$ and here we get the solution $(111,-110)$

$x-y=-221$ implies $x+y=-1$ and here we get the solution $(-111,110)$

Are there any other solutions? If $(x,y)$ is an integer solution, then $x-y$ and $x+y$ would be integer too and in particular they will divide $221$, but we rule out already all the possibilities.

Answer 2

Since $x^2-y^2=(x+y)(x-y)$, putting $s=x+y, d=x-y$ then you must have
$sd=A$ and therefore $\pm s$ \ $A$ and $d=A/s$.

But because $x =(s+d)/2 \, , \; y=(s-d)/2$ shall be integers , then $s,d$ must also have the same parity.

Answer 3

It matters that this is $x^2 - \mathbf{1} y^2 = 221$, so that this is not a Pell equation. Recall that $x^2 - y^2 = (x-y)(x+y)$ since it is the difference of two squares. Since we are requiring that $x$ and $y$ are simultaneously integers, $x-y$ and $x+y$ are simultaneously integers and we know we are seeking all divisors of $221$ in integers, to check which ones give solutions. The divisors of $221$ are $\pm 1$, $\pm 13$, $\pm 17$, and $\pm 221$. Taking these in order as the factor $x+y$, we get the following list. (Since we will end up using each of these as the $x-y$ factor as we go, we do not also need to check cases where we explicitly set $x+y$ to be one of these divisors.) Throughout we use the convention that if $x = \pm \dots$ and $y = \mp \dots$, we use the upper signs for both or the lower signs for both.

• $x+y = \pm 1$, so $x-y = \dfrac{221}{\pm 1} = \pm 221$. Solving this simultaneous system for $x$ by adding the two equations, we have $2x = \pm 222$, so $x = \pm 111$, then $y = \mp 110$.
• $x+y = \pm 13$, so $x-y = \dfrac{221}{\pm 13} = \pm 17$. Again, $2x = \pm 30$ so $x = \pm 15$ and $y = \mp 2$.
• $x + y = \pm 17$, $2x = \pm 30$, $x = \pm 15$, $y = \pm 2$.
• $x + y = \pm 221$, $2x = \pm 222$, $x = \pm 111$, $y = \pm 110$.

This gives all eight solutions.