Showing that a system of Diophantine equations will have irrational solutions as well as integers Solve $\begin{cases} 3xy-2y^2=-2\\ 9x^2+4y^2=10 \end{cases}$ 
Rearranging the 2nd equation to $x^2=\dfrac{10-4y^2}{9} \Longrightarrow 0\leq x^2 \leq 1$ if $x^2=1$ than $y=\pm\dfrac{1}{2}$ and $x=\pm1$ but how do I show there exists two more solutions to this equation by using number theory vs college algebra. Does it have to do with the relationship that $0\leq y^2 \leq 10$ ? So does this say anything about solutions being irrational?
 A: If you add five times the first equation plus the second, you get $9 x^2 + 15 x y - 6 y^2 = 0,$ or
$$ 3 (3x - y)(x+ 2 y) = 0. $$
So the choices lie along either the line $$ y = 3x $$ or the line $$ y = \frac{-x}{2}.  $$
From the ellipse equation $9 x^2 + 4 y^2 = 10,$ we get either $45 x^2 = 10$ and $9 x^2 = 2$ and $(3x)^2 = 2,$ or $10 x^2 = 10$ and $x^2 = 1.$
A: The number of solutions has nothing to do with some of the intersection points being rational, which is a happy coincidence. The way to approach this problem is plain old algebra. Add twice the first equation to the second, and divide through by 3 to get
$$2x y + 3x^2 = 2.$$
Then manipulating the second equation gives
$$9x^4 + 4x^2y^2 = 10x^2$$
$$9x^4 + (2-3x^2)^2 = 10x^2$$
This is a quadratic equation in $x^2$, which has solutions $x^2 = 1, \frac{\sqrt{2}}{3}.$
When $x^2 = 1$, from the second equation we have $y^2 = \frac{1}{4}$, checking these in the first equation reveals that two of the sign choices are spurious, leaving $(-1, \frac{1}{2})$ and $(1, -\frac{1}{2})$ as true solutions.
Similarly, when $x^2 = \frac{\sqrt{2}}{3}$, you get two solutions $(-\frac{\sqrt{2}}{3}, -\sqrt{2})$ and $(\frac{\sqrt{2}}{3}, \sqrt{2})$.
Since your two polynomials have no common polynomial divisor, by Bezout's Theorem, the system 
of polynomial equations has at most four solutions, and we have found them all. (Geometrically, your equations describe a nondegenerate hyperbola and ellipse, which can intersect in at most four points.)
