lqanlf 发表于 2007-3-16 00:34:29

数学高手来看看!!!

<p>根据威尔逊定理,当&nbsp; N!+1&nbsp; 能被&nbsp; N+1&nbsp; 整除时,&nbsp; N+1&nbsp; 就是一个素数.</p><p>设存在这样的一个大于等于零的整数&nbsp; A&nbsp; ,使得对每个自然数N&nbsp;&nbsp;,&nbsp;&nbsp; &nbsp;(N-A)!+1&nbsp;&nbsp;&nbsp; 能被&nbsp; N-A+1 整除(此处A&lt;N),&nbsp;</p><p>那么&nbsp; N-A+1&nbsp;&nbsp; &nbsp;也是一个素数.</p><p>那么,只要证明&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (N+A)!+1&nbsp;&nbsp;&nbsp; 能被&nbsp;&nbsp;&nbsp; N+A&nbsp;+1 &nbsp;整除,那么&nbsp;&nbsp; N+A+1&nbsp; 也是一个素数.</p><p>&nbsp;</p><p>即对于任何&nbsp;&nbsp; N&nbsp;&nbsp; ,都有&nbsp; N-A+1+N+A+1=2N+2&nbsp; </p><p>由于&nbsp;&nbsp; N-A+1&nbsp;&nbsp; 和&nbsp;&nbsp; N+A+1&nbsp;&nbsp; 都是素数,所以由此可以看出,任意大于4的偶数都是两个素数之和.</p><p>只是不知道这个&nbsp;&nbsp;&nbsp; N+A+1&nbsp;&nbsp;&nbsp; 是否真的是一个素数.如果能证明是的话,那么哥德巴赫猜想是否可以说是完成了???</p><p>&nbsp;</p><p>&nbsp;</p><p>这些我都不知道,证明不会,也不知道自己的想法是否是正确的,在这请各位高手指教.</p>

approximat 发表于 2007-3-17 19:24:46

<p>不会得到证明的！</p>

duanyf123 发表于 2007-3-20 05:32:15

<p>首先由于威尔逊定理的<span lang="EN-US"> N </span>值不具有任意性</p><p><span lang="EN-US">&nbsp;</span>所以命题中的<span lang="EN-US"> N-A </span>与<span lang="EN-US"> N+A </span>就不具有任意性</p><p>其次命题本身的叙述也有问题 不可能存在一个<span lang="EN-US"> A </span>对任意<span lang="EN-US"> N </span>值都满足条件</p><p><span lang="EN-US">&nbsp;</span>（因为<span lang="EN-US"> N </span>最小为<span lang="EN-US"> 1 </span>此时<span lang="EN-US"> A==0 </span>若成立 则<span lang="EN-US"> A </span>的唯一取值是<span lang="EN-US"> 0</span>）</p><p><span lang="EN-US">&nbsp;</span>应该表述为对于一个<span lang="EN-US"> N&nbsp;</span>至少存在一个<span lang="EN-US"> A </span>满足条件</p><p>反例：当<span lang="EN-US"> N=3&nbsp;</span>时 对应偶数<span lang="EN-US"> 2N+2=8</span>；</p><p><span lang="EN-US">&nbsp;&nbsp;A </span>的取值范围是<span lang="EN-US"> 0</span>，<span lang="EN-US">1</span>，<span lang="EN-US">2</span>；</p><p><span lang="EN-US">&nbsp;&nbsp;</span>易的只有当<span lang="EN-US">A=2</span>时，<span lang="EN-US">(N-A)!+1=2 </span>才能被<span lang="EN-US"> N-A+1=2 </span>整除【<span lang="EN-US">2</span>为素数】；</p><p><span lang="EN-US">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>而此时<span lang="EN-US">&nbsp; </span>（<span lang="EN-US">N+A)</span>！<span lang="EN-US">+1=121</span>不能被<span lang="EN-US"> N+A+1=6 </span>整除【<span lang="EN-US">6</span>为合数】；</p><p><span lang="EN-US">&nbsp;&nbsp;&nbsp; </span>即<span lang="EN-US"> N=3 </span>时找不到一个<span lang="EN-US"> A </span>使其满足条件；</p><p><span lang="EN-US">&nbsp;&nbsp; </span>所以命题不成立。</p><p><span lang="EN-US">&nbsp;&nbsp; </span>以上意见，仅供参考<span lang="EN-US">~~~</span></p>

lqanlf 发表于 2007-3-22 00:21:23

<div class="quote"><b>以下是引用<i>duanyf123</i>在2007-3-19 21:32:15的发言：</b><br/><p>首先由于威尔逊定理的<span lang="EN-US"> N </span>值不具有任意性</p><p><span lang="EN-US">&nbsp;</span>所以命题中的<span lang="EN-US"> N-A </span>与<span lang="EN-US"> N+A </span>就不具有任意性</p><p>其次命题本身的叙述也有问题 不可能存在一个<span lang="EN-US"> A </span>对任意<span lang="EN-US"> N </span>值都满足条件</p><p><span lang="EN-US">&nbsp;</span>（因为<span lang="EN-US"> N </span>最小为<span lang="EN-US"> 1 </span>此时<span lang="EN-US"> A==0 </span>若成立 则<span lang="EN-US"> A </span>的唯一取值是<span lang="EN-US"> 0</span>）</p><p><span lang="EN-US">&nbsp;</span>应该表述为对于一个<span lang="EN-US"> N&nbsp;</span>至少存在一个<span lang="EN-US"> A </span>满足条件</p><p>反例：当<span lang="EN-US"> N=3&nbsp;</span>时 对应偶数<span lang="EN-US"> 2N+2=8</span>；</p><p><span lang="EN-US">&nbsp;&nbsp;A </span>的取值范围是<span lang="EN-US"> 0</span>，<span lang="EN-US">1</span>，<span lang="EN-US">2</span>；</p><p><span lang="EN-US">&nbsp;&nbsp;</span>易的只有当<span lang="EN-US">A=2</span>时，<span lang="EN-US">(N-A)!+1=2 </span>才能被<span lang="EN-US"> N-A+1=2 </span>整除【<span lang="EN-US">2</span>为素数】；</p><p><span lang="EN-US">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>而此时<span lang="EN-US">&nbsp; </span>（<span lang="EN-US">N+A)</span>！<span lang="EN-US">+1=121</span>不能被<span lang="EN-US"> N+A+1=6 </span>整除【<span lang="EN-US">6</span>为合数】；</p><p><span lang="EN-US">&nbsp;&nbsp;&nbsp; </span>即<span lang="EN-US"> N=3 </span>时找不到一个<span lang="EN-US"> A </span>使其满足条件；</p><p><span lang="EN-US">&nbsp;&nbsp; </span>所以命题不成立。</p><p><span lang="EN-US">&nbsp;&nbsp; </span>以上意见，仅供参考<span lang="EN-US">~~~</span></p></div><p>显然,你是理解错了,</p><p>当N=3时,A=1而不是2,</p><p>当A=0时,N-A+1=3,N+A+1=5</p><p>即两个都是素数.</p><p>这个是很清楚的!</p><p>即4是3与5的等差中项.</p><p></p><p>在这里,我的意思是先<br/>N+1然后再+-A.这样才更容易理解这个意思.</p>

jiwawa 发表于 2007-3-27 15:42:58

<br/>有个问题你没考虑到：（N+A)!+1为奇数，而N+A+1时正时负，因此，(N+A)!+1不一定被N+A+1整除

jiwawa 发表于 2007-3-27 15:45:35

不好意思，是N+A+1时奇时偶

海岩秋沙 发表于 2007-3-29 22:51:00

<p>你想证明哥德巴赫猜想啊！看起来容易，但很难证明的。加油！！</p>
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