# Diophantine solutions to a large geometric figure

I have a related question to one I've read today, see: Integer solutions to $2x^2+5x+y^2=19$

The integers solution are part of an ellipse, with an obvious finite number of $x$. What I would like to know is, when we have a large ellipse so that computing each $x$ would be disastrous by hand, is there any method to cut the possibilities down, or provide an exact answer for an ellipse? Or are there only approximation algorithms that can accomplish this?

If $x$ and $y$ are positive integers you can write: $$5x+y^2\equiv 1\pmod 2$$ But the acceptable values of $x$ are $1,2,3$ because $5x<19$ while the acceptable values of $y$ are $1,2,3,4$ because $y^2<19$. You can consider two systems: $$\left\{ \begin{array}{c} 5x\equiv 0\pmod 2\\ y^2\equiv 1\pmod 2 \end{array} \right.$$ And $$\left\{ \begin{array}{c} 5x\equiv 1\pmod 2\\ y^2\equiv 0\pmod 2 \end{array} \right.$$ In the first systems the acceptable value of $x$ is only $2$ and the acceptable values of $y$ are $1,3$. In the second system the acceptable values of $x$ are $1,3$ while $y$ are $2,4$. Now you can control the solitions. Edit: to find also the negative values you can consider the equation of second degree in $x$: $$2x^2+5x+y^2-19=0$$ The discriminant of equation is $\Delta=177-8y^2$ but $$177-8y^2\ge 0$$ therefore the accetable values of $y$ are:${{-4,-3,....,3,4}}$ but $$177-8y^2=k^2$$ and the only values that solve the equation are $y=$ $\pm 1$ and $\pm 4$.
• Why did you ignore the $2x^2$ and why did you use mod 2? Why only positive integers? – Trademark May 28 '15 at 11:59
• I ignore $x^2$ because I use the $mod 2$ but I could use also $mod 5$ and to ignore $5x$. And I found only positive solutions as I wrote in my answer – Domenico Vuono May 28 '15 at 12:09
• Would finding negative solutions me more complicated? From what you've told me, we can use $2x^2 + y^2 \equiv 4 \pmod 5$, or is it $\equiv 1 \pmod 4$? Would we have $5$ systems for $\pmod 5$? – Trademark May 28 '15 at 12:47
• Using $mod 5$ we obtain the equation $2x^2+y^2\equiv 4\pmod 5$ and we have analyse $5$ systems – Domenico Vuono May 28 '15 at 12:52
• The other four systems are $2x^2\equiv 1, y^2\equiv 3\pmod 5$,$2x^2\equiv 2, y^2\equiv 2\pmod 5$, $2x^2\equiv 3,y^2\equiv 1\pmod 5$ and $2x^2\equiv 0,y^2\equiv 4\pmod 5$ – Domenico Vuono May 28 '15 at 16:25