Explain Hartshorne solution IV.4.6c. Hartshorne IV.4.6c asks:

If $X$ is an elliptic curve, for $d\geq 3$ embed $X$ as a curve of degree $d$ in $\Bbb P^{d-1}$, and conclude that $X$ has exactly $d^2$ points of order $d$ in its group law.

The Solutions PDF says:

$X$ has $d^2$ hyperosculating points.  If X is embedded via $|dP_0|$, then P is a hyperosculating point when it is the divisor of a hyperplane is $dP$ which happens when P has order dividing d in the group law (see also II.6 excercises).

I have no clue what the second sentence in that solution means.  Does anyone have a hint on how to start this problem?
 A: If you choose any point $P_0 \in X$ and embed $X$ into $\mathbb{P}^{d-1}$ by complete linear system of the divisor $dP_0,$ point $P_0$ is a hyperosculating point. Without loss of generality you can take $P_0$ to be the identity in the group $X$.
If $H$ is a hyperplane in $\mathbb{P}^{d-1}$ and intersects with $X$ at a hyperosculating point $P$ then the divisor of intersection is $dP$ on $X$. This divisor is linearly equivalent to the divisor $dP_0$, because all hyperplanes in $\mathbb{P}^{d-1}$ are linearly equivalent. So, $d(P-P_0)=0$, here $0$ is the identity in the divisor class group of $X$. Such point $P$ is a point of order $d$ in $X$. 
Conversely, every point of order $d$ is a hyperosculating point. To see this you use the same argument in the other direction: $P$ has order $d$ consider divisor $dP$ it is linearly equivalent to $dP_0$ so it is a divisor obtained by intersecting $X$ with a hyperplane in $\mathbb{P}^{d-1}$.
Thus, if we know that there are $d^2$ hyperosculating points (this is part b of the exercise) we can conclude that there are $d^2$ points of order $d$. 
