From: Kouhia Juhana Krister <jk87377@cs.tut.fi>
Date: Mon, 17 Apr 1995 15:23:42 +0300
To: rusin@washington.math.niu.edu
Subject: Thanks!

About a month ago you posted to sci.math and transformed
an elliptic curve to its normal form. Using the same
move-one-root-to-infinity method, I succeeded to transform
the y^2 = 2x^4-1 to the normal form very easily.

Thank you for your help, since before I did read your text I were
not exactly sure how to move a root to infinity. Now I know.

I'm writing an algebra class report about the group structure on
elliptic curves and wanted to include the classical Fermat example
which were not done at all in details in the references I have
(exactly: in Alf van der Poorten's FLT lecture notes in e-math.ams.org).

I include the interesting part of the transformation below.

Juhana Kouhia

==clip==
%% Plain TeX

We transform the elliptic curve $y^2 = 2x^4-1$ to the form
$y^2 = x^3 + {1 \over 4}x$ by moving one of the roots of $y^2 = 2x^4-1$
to the infinity and by making a linear change of variables.

Since $2x^4-1 = (\sqrt2 x^2 - 1)(\sqrt2 x^2 + 1) = (\root 4 \of 2 x
- 1)(\root 4 \of 2 x + 1)(\root 4 \of 2 x - i)(\root 4 \of 2 x + i)$,
we substitute $y = v(\root 4 \of 2 x - 1)^2$ and get
$$\eqalignno{v^2 &= {\root 4 \of 2 x - 1 \over \root 4 \of 2 x - 1}
{\root 4 \of 2 x + 1 \over \root 4 \of 2 x - 1}{\root 4 \of 2 x - i
\over \root 4 \of 2 x - 1}{\root 4 \of 2 x + i \over \root 4 \of 2 x - 1}
\cr
\noalign{\vskip 5pt}
&= \left(1 + {2 \over \root 4 \of 2 x - 1}\right) \left(1 + {1 - i \over
\root 4 \of 2 x - 1}\right) \left(1 + {1 + i \over \root 4 \of 2 x - 1}\right)
\cr
\noalign{\vskip 5pt}
&= (1 + 2u)(1+(1-i)u)(1+(1+i)u) \cr
\noalign{\vskip 5pt}
&= 4u^3 + 6u^2 + 4u + 1, \cr}$$
where $u = {1 \over \root 4 \of 2 x - 1}$.

\bye