Can adding two rational points on an elliptic curve yield an integral point?

Question

Let $E : y^2 = x^3 + ax + b$ be an elliptic curve with $a,b \in \Bbb{Z}$. We know that by the group law, the set of rational points in $E$ forms an abelian group with $+$ as in defined in the link above.

Let $P, Q \in E$ be two points in $E$ such that one of them has rational but non-integral coordinates. Is it possible that $P + Q$ has integral coordinates? What if we restrict it to $P + P$, where $P$ has coordinates in $\Bbb{Q} \setminus \Bbb{Z}$? Trying out some examples seems to show that $P + Q$ can't have integral coordinates, but I can't seem to prove it using elementary methods.

Context: This problem is related to an attempt to show that an integral point $P$ is not torsion if $P + P$ does not have integral coordinates (not sure if this is even true). If the statement above holds, then this follows easily. Perhaps there is a much easier way to do this?

Answer 1

Take any integral point $P$ and any rational non-integral point $Q$ such that $P-Q$ is not integral ( these are easy to find since the set of integrl points is finite). Then $P=Q+(P-Q)$.

Answer 2

Reiterating my comment from above, take $Q = -P + R$, where $R$ is any integral point for counterexamples to the first question.

For the case of $P + P$, then the answer is yes. Take $E: y^2 = x^3 + \frac{1}{4} x$. Then $E(\mathbf{Q})_\mathrm{tor} \cong \mathbf{Z}/4\mathbf{Z}$ and is generated by $P = \left(\frac{1}{2},\frac{1}{2} \right) \in E(\mathbf{Q}\setminus \mathbf{Z})$, but $2P = (0, 0)$.

Answer 3

Here’s an easy way to see that if $P$ has nonintegral coordinates, then $[2](P)$ may well have integral coordinates:

I’ll show that the contrapositive of your proposed proposition is false. This says: if $Q$ has coordinates in $\Bbb Z$, then all four $P$ with $[2](P)=Q$ have coordinates in $\Bbb Z$. But if that were true, then there would be infinitely many points with $\Bbb Z$-coefficients, well known to be false.

(I recognize that the above argument is not self-contained, needing as it does support from fairly advanced theorems on elliptic curves. I’ll let others fill in the details, if they’re so inclined.)

(Note to Edit): I thank @ user760870 for making clear that my use of the word “integral” was imprecise.