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楼主 |
发表于 2009-12-11 17:58
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一句话证明费马大定理成立
````追寻费马笔录真谛六探——“神州的智附愚绅传唱、谀附者高唱、盲附者大唱”的部份名著摘录↓
````1979年版华罗庚《数论导引》318页定理1。无整数能适合X^4+y^4= Z^2。……;
````1980年中文版u•杜德得《基础数论》148页定理1 下列方程没有非平凡解:X^4+y^4= Z^2。
``````````````````注意,这也意味着X^4+y^4= Z^4无解,因为若a,b,c是此方程的解,则有a^4+b^4=( c^2) ^2,````````````与定理1矛盾。……;
````1979年版夏圣亭《不定方程浅说》165页 例65 证明 方程X^4+y^4= Z^4无正整数解。
`````````````````````````````````````````````````````````````````````证明:以u= Z^2代替Z^4,得方程
``````````````````````````````````````````````````````````````````````````` X^4+y^4= Z^2。……; (1)
````1979年版柯召 孙琦《谈谈不定方程》92页(本楼主将重点以此版本来介绍其中的伪迹):
````为了证明费马大定理,实际上只须证明不定方程X^4+y^4= Z^4和不定方程X^p+y^p= Z^p,p是奇素数均无xyz≠0的整数解。……。
````以上这些抄袭“递降法”的名著,全都嫁祸并践踏费马的好名声,把忽悠人的谬论伪装成“高级真理”:
````由(1)设定X^4+y^4= Z^2为最小,则据欧氏二奇一偶勾股数构造公式又有
````````````````````````````````````````````````````````````````````x^2=a^2-b^2, y^2=2ab, z=a^2+b^2。 (2)
而由(2)x^2=a^2-b^2 → x^2+b^2=a^2 又得
```````````````````````````````````````````````````````````````````````x=p^2-q^2, b=2pq, a=p^2+q^2。 (3)
从而仍由(2)y^2=2ab与(3)b=2pq,a=p^2+q^2又得
`````````````````````````````````````````````````````````````````````````````````y^2=2ab=4pq(p^2+q^2)。 (4)
据(4)写p=r ^2,q=s^2, p^2+q^2=Z1^2则又获得与(1)同模但比(1)更小的解为↓
```````````````````````````````````````````````````````````````````````````````r^4+s^4= Z1^2。 (5)
(5)与(1)同模但其解Z1小于Z ,有悖于Z是最小给定数之假设。证明(1)无正整数解。
````以上伪迹概括有3,
1,无任何根据而臆定(1)为给定最小数,是数学的唯心主义;
2,循环论证;
3,(1)的写法暗地里把底数的充分条件为Z>X、Z>Y、X+Y>Z最重要的“X+Y>Z”砍掉了。待续。
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