Here are two points,P (-1,4), P (2,5)
Which can be easily verified by substituting them into E.
Found some more points, (-2,3), (4,9), (8,23), (43,282), (52,375) So we have a total of 7 points.
Maybe someone could lend a hand by implementing
for i in range(-3,n):
x = sqrt(i^3 + 17)
if x == int
print i,x
To see if there are any more values, (I tried to put this code into SAGE but doesn't seem to work, could anyway tell me why not, and how I could get it working?) I will ask my computer science friend if not =)
Cheers guys
Oooh I skimmed past the first 10000 points, must of missed that one, good job. Did you write a program for that? The gap seems to be getting very large between the points, will be interesting to how many more there are! Any idea how to write a program in SAGE that would find such points?
Thanks for taking an interest!
No, I just used the fact that if you have two integer points on the curve such that the line joining them has integer-valued gradient, then that line will meet the curve at a third such point. The line joining (43,–282) and (52,375) has gradient 73, and meets the curve again at (5234,378661).
That is a neat technique for finding such points, but of course it is not going to help in proving that there are only finitely many of them.
Edit. I just came across this reference to a theorem of Siegel. It also states that the eight integer solutions to y^2=x^3+17 are the only ones with y>0.