 Solution

At $P_1$ the tangent slope is zero. The elliptic-curve addition formula givesThe chord from $P_1$ to $P_2$ has slope $(-3-15)/(-6)=3$, so its sum has $x$-coordinate $3^2-0-(-6)=15$ and $y$-coordinate $3(0-15)-15=-60$. Thus $P_1+P_2=(15,-60)$.

Reduction at the good prime seven gives $\tilde{E}:y^2=x^3+1$. For $x=0,1,2,3,4,5,6$, the right sides are respectively $1,2,2,0,2,0,0$. There are respectively $2,2,2,1,2,1,1$ choices of $y$. Including the point at infinity gives $\#\tilde{E}({\mathbb{F}}_7)=12$. Its nonzero torsion points of an elliptic curve of order two are $(3,0),(5,0),(6,0)$, so it cannot be cyclic. Its two-primary component is $(\mathbb{Z}/2\mathbb{Z}{)}^2$ and its three-primary component is cyclic of order three; henceIn particular its exponent is six.

The reduction of an elliptic curve is a group homomorphism defined on every ${\mathbb{Q}}_7$-point by projectivity. Thus $6P$ reduces to the identity for every rational $P$. A finite point with integral coordinates reduces to an affine point, whose projective last coordinate is one, so it cannot reduce to the identity at infinity. Therefore every nonzero finite point $6P$ has nonintegral coordinates. If $6P=O$, it has no affine coordinates at all. This is the reduction exponent obstruction to integral multiples.