Fermat once claimed that the only integral solutions to y2=x3−2 are (3,pm5).
Fermat knew Bachet's duplication formulas (more precisely, Bachet had a formula for computing what we call −2P), which for y2=x3+ax+b says
x2P=fracx4−8bx4x3+4b=fracx4cdotfracx3−8bx3+b.
Using this formula it is easy to prove the following:
Consider the point P=(3,5) on the elliptic curve y2=x3−2.
The x-coordinate xn of [−2]nP has a denominator divisible by
4n; in particular, [−2]nP has integral coordinates only if n=0.
In fact, writing xn=pn/qn for coprime integers pn, qn, we find
xn+1=fracxn4cdotfracx3n+16x3n−2=fracpn4qncdotfracp3n+16q3np3n−2q3n.
Since pn is odd for nge1 and qn=4nu for some odd number u (use induction), we deduce that the power of 2 dividing qn+1 is 4 times that dividing qn.
My question is whether the general result that kP has integral affine coordinates if and only if k=pm1 can be proved along similarly simple lines. The modern proofs based on the group law, if I recall it correctly, use Baker's theorem on linear forms in logarithms.
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