2 proofs Regarding Elliptic curves

I am self studying Elliptic curves / Algebra from book "Elliptic Curve Number theory And Cryptography second edition" I got stuck trying to do these 2 proofs questions.

Let $n$ be an integer. Show that if $x, y$ are rational numbers satisfying $y^ 2 = x^3 − n^2x$, and $x \neq 0, ±n$, then the tangent line to this curve at $(x, y)$ intersects the curve in a point $(x_1 , y_1 )$ such that $x_1 , x_1 − n, x_1 + n$ are squares of rational numbers.

Actually this question got a first part that was easy to answer:

Show that if $x, y$ are rational numbers satisfying $y^2 = x^3 − 25x$ and $x$ is a square of a rational number, then this does not imply that $x + 5$ and $x − 5$ are squares.

It was easy to proof by 1 counter example $x = 25/4$ which was already given as a hint.

The second question

Let $(x, y)$ be a point on the elliptic curve E given by $y^2 = x^3 + Ax + B$. Show that if $y = 0$ then $3x^2 + A \neq 0$. (Hint: What is the condition for a polynomial to have x as a multiple root?)

Besides the solution I would appreciate any tips on how to think about doing proofs in general in ECC or Abstract algebra.

• ops that was the question number, and the qual should be not equal EDIT: Fixed – oddcoder Mar 10 '18 at 16:47
• It is always a good idea to give the book's author. I think you mean Washington's book, and a hint for you in 2.5: a polynomial has a multiple root $\;\alpha\;$ iff its derivative polynomial also has $\;\alpha\;$ as one of its roots. – DonAntonio Mar 10 '18 at 16:48
• I did fix the typo right now, Equal should have been Not Equal, sorry for the mistake – oddcoder Mar 10 '18 at 16:51
• In the first question, do you mean $x\ne0$, $\pm n$? – Lord Shark the Unknown Mar 10 '18 at 16:52
• It seams that I did lots of mistakes while typing let me fix it and recheck – oddcoder Mar 10 '18 at 16:55

The tangent line to $E$ at a point $(x_0,y_0)$ usually has equation $$(y-y_0)=m(x-x_0)\tag{1}$$ where $m$ is the value of $dy/dx$ at the point $(x_0,y_0)$. From $$y^2=x^3-n^2x\tag{2}$$ we get $$2y\frac{dy}{dx}=3x^2-n^2,$$ that is $$m=\frac{3x_0^2-n^2}{2y_0}.$$ To find the intersection points of $E$ with the tangent, putting $(1)$ into $(2)$ gives $$x^3-m^2x^2+(\cdots)x+(\cdots)=0.\tag{3}$$ Let's not worry what the coefficients of $x^1$ and $x^0$ are. The sum of the three roots of $(3)$ are $m^2$. But these three roots are $x_0$, $x_0$ and $x_1$. The root $x_0$ is repeated as the line is a tangent to the curve there. So $$x_1=m^2-2x_0=\frac{(3x_0^2-n^2)^2}{4y_0^2}-2x_0 =\frac{(3x_0^2-n^2)^2}{4(x_0^3-n^2x_0)}-2x_0$$ etc. Now you have to massage $x_1$ and $x_1\pm n$ into a form that makes it clear that all of them are square...
• there i kind of text alignment problem in your third statement $x^3 - m^2x^2....$ – oddcoder Mar 11 '18 at 9:23
$$F(x,y)=y^2-x^3-Ax-B=0\Rightarrow \begin{cases}\dfrac{\partial F(x,y)}{\partial y}=2y\\\dfrac{\partial F(x,y)}{\partial x}=-(3x^2+A)\end{cases}$$ It follows that if $y=0$ then $\dfrac{\partial F(x,y)}{\partial y}=0$ so if $3x^2+A=0$ then the curve would be singular (i.e. non-elliptic) since $\dfrac{\partial F(x,y)}{\partial x}$ would be null also.
Consequently $3x^2+A\ne 0$
• can you please please elaborate what theorem, law or rule is the thing that implies that if $y=0$ then partial derivative should also be equal to zero ? – oddcoder Mar 10 '18 at 21:00
• When the elliptic cubic is taken as its Weirstrass form you have $y^2+f(x)=0$ so the partial derivative is $2y$. Tha's all, it is just an evident fact. – Piquito Mar 10 '18 at 23:43