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Frey Curve

Let $a^p+b^p=c^p$ be a solution to Fermat's Last Theorem. Then the corresponding Frey curve is

\begin{displaymath}
y^2=x(x-a^p)(x+b^p).
\end{displaymath} (1)

Frey showed that such curves cannot be Modular, so if the Taniyama-Shimura Conjecture were true, Frey curves couldn't exist and Fermat's Last Theorem would follow with $b$ Even and $a\equiv -1\ \left({{\rm mod\ } {4}}\right)$. Frey curves are Semistable. Invariants include the Discriminant
\begin{displaymath}
(a^p-0)^2(-b^p-0)[a^p-(-b)^p]^2 = a^{2p}b^{2p}c^{2p}.
\end{displaymath} (2)

The Minimal Discriminant is
\begin{displaymath}
\Delta=2^{-8} a^{2p}b^{2p}c^{2p},
\end{displaymath} (3)

the Conductor is
\begin{displaymath}
N=\prod_{l\vert abc} l,
\end{displaymath} (4)

and the j-Invariant is
\begin{displaymath}
j={2^8(a^{2p}+b^{2p}+a^p b^p)^3\over a^{2p}b^{2p}c^{2p}}={2^8(c^{2p}-b^pc^p)^3\over(abc)^{2p}}.
\end{displaymath} (5)

See also Elliptic Curve, Fermat's Last Theorem, Taniyama-Shimura Conjecture


References

Cox, D. A. ``Introduction to Fermat's Last Theorem.'' Amer. Math. Monthly 101, 3-14, 1994.

Gouvêa, F. Q. ``A Marvelous Proof.'' Amer. Math. Monthly 101, 203-222, 1994.




© 1996-9 Eric W. Weisstein
1999-05-26