= 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 $p$,
$$
v_p(x)=-2r,\qquad v_p(y)=-3r,\qquad r>0.
$$
Thus both coordinates of $5P$ have negative two-adic <valuation>, and both coordinates of $7P$ have negative three-adic <valuation>. In particular
$$
\boxed{x(5P),y(5P),x(7P),y(7P)\notin\mathbb Z.}
$$
There 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 gives
$$
4P=(2,-3),\qquad5P=(1/4,-5/8),\qquad
6P=(6,14),\qquad7P=(-5/9,8/27).
$$
The original printed part repeats its request about $7P$ at the end; the two distinct requested multiples are $5P$ and $7P$.
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