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	<title>Carmichael number is not semiprime - Revision history</title>
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		<title>Vipul: Created page with &quot;==Statement==  Suppose &lt;math&gt;n&lt;/math&gt; is a composite natural number that is a Carmichael number, i.e., it is a Fermat pseudoprime to every base relatively prime to...&quot;</title>
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		<updated>2012-01-15T21:14:34Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot;==Statement==  Suppose &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; is a composite &lt;a href=&quot;/w/index.php?title=Natural_number&amp;amp;action=edit&amp;amp;redlink=1&quot; class=&quot;new&quot; title=&quot;Natural number (page does not exist)&quot;&gt;natural number&lt;/a&gt; that is a &lt;a href=&quot;/wiki/Carmichael_number&quot; title=&quot;Carmichael number&quot;&gt;Carmichael number&lt;/a&gt;, i.e., it is a &lt;a href=&quot;/wiki/Fermat_pseudoprime&quot; title=&quot;Fermat pseudoprime&quot;&gt;Fermat pseudoprime&lt;/a&gt; to every base relatively prime to...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;==Statement==&lt;br /&gt;
&lt;br /&gt;
Suppose &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; is a composite [[natural number]] that is a [[Carmichael number]], i.e., it is a [[Fermat pseudoprime]] to every base relatively prime to it. Equivalently, the [[universal exponent]] &amp;lt;math&amp;gt;\lambda(n)&amp;lt;/math&amp;gt; divides &amp;lt;math&amp;gt;n- 1&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Then, &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; is not a [[semiprime]].&lt;br /&gt;
&lt;br /&gt;
==Facts used==&lt;br /&gt;
&lt;br /&gt;
# [[uses::Carmichael number is square-free]]&lt;br /&gt;
&lt;br /&gt;
==Proof==&lt;br /&gt;
&lt;br /&gt;
We prove the contrapositive: a semiprime cannot be a Carmichael number.&lt;br /&gt;
&lt;br /&gt;
By Fact (1), it suffices to restrict attention to a Carmichael number of the form &amp;lt;math&amp;gt;pq&amp;lt;/matH&amp;gt; where &amp;lt;math&amp;gt;p,q&amp;lt;/math&amp;gt; are distinct primes.&lt;br /&gt;
&lt;br /&gt;
We have:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\lambda(n) = \operatorname{lcm} \{ p - 1, q - 1\}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
In particular, for &amp;lt;math&amp;gt;\lambda(n)&amp;lt;/math&amp;gt; to divide &amp;lt;math&amp;gt;n - 1&amp;lt;/math&amp;gt;, we must have:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;p - 1 \mid n - 1 \qquad \mbox{ and } \qquad q - 1 \mid n - 1&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The first condition tells us that:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;n \equiv 1 \pmod{p - 1}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since we already have &amp;lt;math&amp;gt;p \equiv 1 \pmod{p - 1}&amp;lt;/math&amp;gt;, and since &amp;lt;math&amp;gt;n = pq&amp;lt;/math&amp;gt;, this gives &amp;lt;math&amp;gt;q \equiv 1 \pmod{p - 1}&amp;lt;/math&amp;gt;. In particular, this gives &amp;lt;math&amp;gt;q \ge p&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Similarly, the second condition tells us that &amp;lt;math&amp;gt;p \equiv 1 \pmod{q - 1}&amp;lt;/math&amp;gt;. In particular, this gives &amp;lt;math&amp;gt;p \ge q&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Combining, we get &amp;lt;math&amp;gt;p = q&amp;lt;/math&amp;gt;, contradicting our requirement that &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; be a product of distinct primes.&lt;/div&gt;</summary>
		<author><name>Vipul</name></author>
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