费马大定理的证明方法是无限的，有两个推论
Fermat's Last Theorem admits an infinity of proving ways and two corollaries
摘要：
费马的陈述等价于说，如果$x$，$y$，$z$，$n$是整数，$n>2$，那么$z^{n}\gtrless x^{n}+y^{n}$。这是证明的援助数字$λ$的形式$λ=z/ρ$，与$1<ρ<z$，命名为\emph{reversors}在文本中，因为他们的财产乘以$z^{n-1}$在$z^{n-1}<x^{n-1}+y^{n-1}$，不仅扭转了信号，但也给$z^{n}>x^{n}+y^{n}$作为一个解决方案的逆转不等式。当$λ^{\prime }s$满足相容的反向不等式系统时，$λ$ -集等价于$\mathbb{R}^{+}$区间上的点。因此，定理承认一个不可数无穷的证明方法，每一个由$λ$的一个特定值给出。在推论1中，导出了$y$、$x$、$z$和$n$之间的一般关系。推论2表明，费马陈述中的丢番图方程除了代数无理数和固有的复形之外不存在解。三联体可以分为七组，在每一组中，它们与各自的$n$的关系与表1所示相同。用上述七组算例进行了数值验证，结果与理论完全一致。
英文摘要：
Fermat's statement is equivalent to say that if $x$, $y$, $z$, $n$ are integers and $n>2$, then $z^{n}\gtrless x^{n}+y^{n}$. This is proved with the aid of numbers $λ$'s, of the form $λ=z/ρ$, with $1<ρ<z$, named \emph{reversors} in the text, because their property of multiplying $z^{n-1}$ in $z^{n-1}<x^{n-1}+y^{n-1}$, not only reverses the signal but also gives $z^{n}>x^{n}+y^{n}$ as a solution of the reversed inequality. As the $λ^{\prime }s$ satisfy a compatible opposed sense system of inequalities, the $λ$-set is equivalent to the points of an $\mathbb{R}^{+}$ interval. Therefore the theorem admits a noncountable infinity of proving ways, each one given by a particular value of $λ$. In Corollary 1 a general relation between $y$, $x$, $z$ and $n$ is derived. Corollary 2 shows that the Diophantine equation in Fermat's statement admits no solutions other than algebraic irrationals and the inherent complexes. Integer triplets can be classified in seven sets, within each one their relation with the respective $n$ is the same as shown in Table 1. Numerical verification with examples taken from all the mentioned seven sets gives a total agreement with the theory.
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