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#1
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ÊÁ¡ÒÃä´âÍΌ䷹ì
¨§áÊ´§ÇèÒÊÁ¡ÒÃ
$\frac{1}{X^4}$-$\frac{1}{Y^4} $ =$\frac{1}{Z^4} $ äÁèÁդӵͺ·Õèà»ç¹¨Ó¹Ç¹àµçÁºÇ¡ |
#2
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äÁèÁդӵͺà»ç¹¨Ó¹Ç¹àµçÁºÇ¡
$ x^4+y^4 = k^2 $ |
#3
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ÍéÒ§ÍÔ§:
»Å. ãªé contradiction à¢Õ¹ formal proof ¡çä´é¹Ð |
#4
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ä˹æ¡çà¢éÒÁÒáÅéÇ ...
Lemma $x^2=y^4+z^4$ äÁèÁÕ integers solution ãËé $S$ à»ç¹à«µ¤ÓµÍº¢Í§ÊÁ¡Òà $\frac{1}{x^4}-\frac{1}{y^4}=\frac{1}{z^4}$ ãËé $T$ à»ç¹à«µ¤ÓµÍº¢Í§ÊÁ¡Òà $x^2=y^4+z^4$ ¨Ò¡ Lemma ¨Ðä´éÇèÒ $T$ à»ç¹à«µÇèÒ§ à»ç¹¡ÒÃà¾Õ§¾Í·Õè¨Ð¾ÔÊÙ¨¹ìÇèÒ $S$ à»ç¹à«µÇèÒ§´éÇ ÊÁÁµÔãËé $S$ äÁèà»ç¹à«µÇèÒ§ ´Ñ§¹Ñ鹨ÐÁÕ $(x_{0},y_{0},z_{0}) \in S$ ´Ñ§¹Ñ鹨Ðä´éÇèÒ $\frac{1}{x_{0}^4}-\frac{1}{y_{0}^4}=\frac{1}{z_{0}^4}$ ¨Ñ´ÃÙ»à»ç¹ $(y_{0}^2z_{0}^2)^2=(z_{0}x_{0})^4+(x_{0}y_{0})^4$ ---(*) ãËé $(a,b,c)=(y_{0}^2z_{0}^2,z_{0}x_{0},x_{0}y_{0})$ à¾ÃÒÐÇèÒ $x_{0},y_{0},z_{0}$ à»ç¹¨Ó¹Ç¹àµçÁ ¨Ðä´é $a,b,c$ à»ç¹¨Ó¹Ç¹àµçÁ´éÇ áÅШҡ (*) ¨Ðä´éÇèÒ $a^2=b^4+c^4$ ÁÕ $(a,b,c)=(y_{0}^2z_{0}^2,z_{0}x_{0},x_{0}y_{0})$ à»ç¹¤ÓµÍº ´Ñ§¹Ñé¹ $(a,b,c) \in T$ ¢Ñ´áÂ駡Ѻ¡Ò÷Õè $T$ à»ç¹à«µÇèÒ§ ´Ñ§¹Ñé¹ $S$ à»ç¹à«µÇèÒ§ ...QED... »Å. Proof ¢Í§ Lemma google ä´é¹Ð¤ÃѺ ãªé¤ÓÇèÒ infinite descent |
#5
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ÍéÒ§ÍÔ§:
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