Could the solutions of an elliptic curve generate the primes? Take an elliptic curve over $\mathbb{Q}$:
$$ y^2=x^3+ax+b$$
For some $a$ and $b$. 
Let $(x_i,y_i) $ be the $i$th solution to this equation. Is it possible that for sufficient $a$ and $b$, $x_i$ could be the $i$th prime number? The $y$ values can be any rational number.
Obviously I am not asking for explicit $a$ and $b$, but is there any result or theorem that disallows this?
 A: Yes, this is impossible.  I'll only discuss the case where $a$ and $b$ are rational: I'm pretty sure the irrational case reduces to two simultaneous elliptic curves by $\mathbb Q$-linearity.
Clearly if $a$ and $b$ are both integers, then this follows immediately from Siegel's theorem that an elliptic curve has finitely many points, since $x^3 + ax + b$ is an integer and its square root (if rational) must be integral.
Now suppose $a$ and $b$ are rational, with least common denominator $d$ (so that $ad, bd\in \mathbb Z$).  If $(x_0,y_0)$ is a rational point on $y^2 = x^3 + ax + b$ with $x_0\in\mathbb Z$, then 
$(d^2 x_0, d^3 y_0)$ is an integer point on the elliptic curve $y^2 = x^3 + d^4a x + d^6b$, which has integer coefficients.  So again there are only finitely many cases of the latter.
But let's suppose we want to broaden the question a bit, and allow $x$ to encode the primes in some other way (such as $x_i = 1/p_i$).  Then there is still an argument to be made against this possibility.  The rational points on an elliptic curve $E$ have the structure of a finitely-generated abelian group, whose elements grow roughly exponentially in height with each generator.  The number of rational points of height $\le H$ is thus pretty small, only $O(\log^k H)$ where $k$ and the implied constant depend on the rank of $E$ and the size of the torsion group.
Since the number of primes up to $H$ is much, much greater, this is enough to rule out many reasonable potential encodings of the primes into $x$.  But I don't know enough to really say it is completely impossible (I suspect it's true that "$x_i$ has denominator $p_i$" is impossible but I don't know a proof that excludes it).
