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	<title>Wolstenholme&#039;s theorem - Revision history</title>
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	<updated>2026-06-06T06:43:52Z</updated>
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		<title>Vipul: Created page with &quot;==Statement==  ===Statement in terms of near-central binomial coefficients===  Suppose &lt;math&gt;p&lt;/math&gt; is a prime number greater than 3. Then, we have the following congrue...&quot;</title>
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		<updated>2012-01-03T20:35:50Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot;==Statement==  ===Statement in terms of near-central binomial coefficients===  Suppose &amp;lt;math&amp;gt;p&amp;lt;/math&amp;gt; is a &lt;a href=&quot;/wiki/Prime_number&quot; title=&quot;Prime number&quot;&gt;prime number&lt;/a&gt; greater than 3. Then, we have the following congrue...&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;
===Statement in terms of near-central binomial coefficients===&lt;br /&gt;
&lt;br /&gt;
Suppose &amp;lt;math&amp;gt;p&amp;lt;/math&amp;gt; is a [[prime number]] greater than 3. Then, we have the following congruence condition on the [[fact about::binomial coefficient]]:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\binom{2p - 1}{p - 1} \equiv 1 \pmod {p^3}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Statement in terms of prime cancellation from binomial coefficient===&lt;br /&gt;
&lt;br /&gt;
Suppose &amp;lt;math&amp;gt;p&amp;lt;/math&amp;gt; is a [[prime number]] greater than 3. Then, we have the following congruence condition:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\binom{ap}{bp} \equiv \binom{a}{b} \pmod{p^3}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This version is due to Glaisher and is similar in structure to [[Lucas&amp;#039; theorem]].&lt;br /&gt;
&lt;br /&gt;
===Statement in terms of generalized harmonic numbers===&lt;br /&gt;
&lt;br /&gt;
The following two congruences hold:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\! 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{p-1} \equiv 0 \pmod{p^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
and&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\! 1 + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{(p-1)^2} \equiv 0 \pmod{p}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
What we mean by this is that if we simplify the expressions on the left side and bring them into fractions in reduced form, the numerators are divisible by &amp;lt;math&amp;gt;p^2&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;p&amp;lt;/math&amp;gt; respectively.&lt;/div&gt;</summary>
		<author><name>Vipul</name></author>
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