Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 32 / 1 / v / Solution

The good-reduction map respects elliptic addition. Since the finite groups at two and three have orders five and seven, respectively, $5P$ reduces to $O$ at two and $7P$ reduces to $O$ at three. These multiples are not $O$, since $P$ has infinite order. The formal kernel of a minimal Weierstrass equation gives, for every nonidentity point in the reduction kernel at $p$,Thus both coordinates of $5P$ have negative two-adic valuation, and both coordinates of $7P$ have negative three-adic valuation. In particularThere is also an elementary way to see that both coordinates must fail integrality here. An integral affine pair cannot reduce to the point at infinity. If one coordinate had negative valuation and the other nonnegative valuation, the equation would give different valuations on its two sides. When both are negative the lowest-valuation terms give $2v_p(y)=3v_p(x)$, so both are indeed negative.

For a direct numerical confirmation, continuing the chord-and-tangent group law calculation givesThe original printed part repeats its request about $7P$ at the end; the two distinct requested multiples are $5P$ and $7P$.