Just started studying eliptic curves and am having trouble with this question. An explanation/solution would be much appreciated.
Find the order of the point X on the elliptic curve $E/Q$ for the two cases:
$X = (0,16)$ on $y^2 = x^3 + 256$
$X = (0.5,0.5)$ on $y^2 = x^3 + 0.25x$
I will describe the group law for an elliptic curve in short form, ie $E: y^2=x^3+Ax+B$ for some $A$, $B$.
The group law on the elliptic curve can be defined geometrically, if $P$, $Q$ are distinct points, then draw the straight line between the two points. This line will intersect the curve with multiplicity $3$ so let $R$ be the third point. Finally if $R=(x,y)$, then reflect in the $x$-axis and take $P+Q=(x,-y)$ and this defines an abelian group. Note here that $R=(x,y)$ is the inverse of $P+Q$, i.e. $(P+Q)+R=O$, where $O$ acts as the identity and is the point at infinity.
There are a couple of exceptions to this:
$1)$ If $P=Q$, then we take the tangent line to the curve instead to find $R$.
$2)$ If either $P=O$ (respectively $Q=O$), we set $P+Q=Q$ (respectively $P$).
$3)$ If $P=(x,y)$ and $Q=(x,-y)$, then the line is vertical and we set $R=P+Q=O$.
Using this we note that if $P \in E$ has the form $P=(x,0)$ then by point $3)$, $P=-P$, ie $2P=P+P=O$.
Now let's calculate the order of your points.
Let $E: y^2=x^3+256$, $X=(0,16)$.
We obtain the derivative of the curve which is $2y \dfrac{dy}{dx} = 3x^2$ so at $X$, we have $\dfrac{dy}{dx}=0$ so the tangent line has gradient $0$. Further the tangent passes through $X$ so the equation is $y=16$. Substituting this back into our original equation, we get $16^2=x^3-256$ so $x^3=0$.
The only solutions to this is $x=0$, so our point $R$ is again $X=(0,16)$. Hence $2X=X+X=(0,-16)$ by above. Now notice that $X=-2X$ so in particular $3X=O$ and since $X \neq O$, the order of $X$ must be $3$.
Now your second curve is $E: y^2=x^3+x/4$ with $X=(1/2,1/2)$. Again the derivative of $E$ is $2y \dfrac{dy}{dx} = 3x^2+1/4$ so the gradient at $X$ is $1$ and so the equation of the tangent is $y-1/2=x-1/2$, ie $y=x$.
Putting this back into $E$, we get $x^2=x^3+x/4 = x(x-1/2)^2$. Since $X$ has $x$ coordinate $1/2$, $R$ must have $x$ coordinate $0$. Putting this back into our equation for the tangent line $y=x$ we get $R=(0,0)$ and hence $2X=X+X=(0,0)$.
Now we again apply point $3)$, since the $y$-coordinate is $O$ to find that $4X=O$ so $X$ has order at most $4$. However, by our calculations $2X \neq 0$ so it cannot have order $2$ and hence the order equals $4$.