TITLE: How prove $ y^2=x^3+x+1370^{1370}$ has at least 6 answers in $ \mathbb{Q}$?
QUESTION [7 upvotes]: How prove that $  y^2=x^3+x+1370^{1370}$ has at least 6 answers in $ \mathbb{Q}$?

REPLY [7 votes]: Here is how I would start:
First we set $L := 1370^{685}$ for readability.
Two trivial solutions for $y^2=x^3+x+L^2$ are $P_{1,2}=(0,\pm L)$.
Using the double formula for elliptic curves given here we get two new points on the curve: 
$\lambda = \frac{1}{2L} ; x_3=\frac{1}{4L^2} ; y_3=-(L+\frac{1}{8L^3})$
Plugging it in it can be veryfied that $P_{3,4}=(x_3,\pm y_4)$ are indeed on the curve.
My guess would be that another doubling or maybe an addition will lead to another two solutions. I don't have the time right now to do it. I might come back later and test it.
Edit:
5th and 6th solution by adding $P_1$ and $P_3$:
$P_{5,6}=(\pm 8L^2(8L^4+1) , 512L^9+96L^5+3L)$
@Nate: Good point.