= Solution
At $P_1$ the tangent slope is zero. The <elliptic-curve addition formula> gives
$$
\boxed{2P_1=(0,-15)=-P_1.}
$$
The 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 \b[$P_1+P_2=(15,-60)$].
Reduction at the good prime seven gives $\widetilde E:y^2=x^3+1$. 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 \b[$\#\widetilde E(\mathbb F_7)=12$]. 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; hence
$$
\boxed{\widetilde E(\mathbb F_7)\cong\mathbb Z/2\mathbb Z\times\mathbb Z/6\mathbb Z.}
$$
In 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 \b[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>.
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