 Solution

Double $2Q=(12,52)$ using the tangent formula. The slope issoThis finite point is neither $P$ nor $-P$, since its abscissa is nonzero. Therefore $P\pm 4Q$ are both finite nonidentity points.

For a direct exact-coordinate proof, the line from $P$ to $4Q$ has slope $985/104$, while the line from $P$ to $-4Q=(264/169,-3944/2197)$ has slope $-493/429$. For a secant through $P=(0,0)$ and $R=(u,v)$, its slope is $m=v/u$, and the group law gives $x(P+R)=m^2-1-u$, $y(P+R)=13-mx(P+R)$. Simplification yieldsThe first abscissa has odd numerator and denominator $64$. The second has denominator $1089=3^2\cdot 11^2$ and numerator not divisible by either $3$ or $11$. Thus neither point has integral coordinates.

Reduction also explains the obstruction without requiring the full secant arithmetic. At the good prime $2$, $\overline{Q}=(0,1)=-\overline{P}$, and $\overline{P}$ has order three, so $\overline{P+4Q}=O$. At the good prime $11$, the displayed coordinates of $4Q$ reduce to $(0,0)=\overline{P}$, so $\overline{P-4Q}=O$. A finite point with integral affine coordinates reduces to an affine point, not to the identity at infinity. Membership in the kernel of reduction of an elliptic curve therefore certifies the same two nonintegrality conclusions, at primes $2$ and $11$ respectively.