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	<id>https://charlesreid1.com/w/index.php?action=history&amp;feed=atom&amp;title=FMM10</id>
	<title>FMM10 - Revision history</title>
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	<updated>2026-06-19T10:48:35Z</updated>
	<subtitle>Revision history for this page on the wiki</subtitle>
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	<entry>
		<id>https://charlesreid1.com/w/index.php?title=FMM10&amp;diff=27043&amp;oldid=prev</id>
		<title>Admin: Created page with &quot;{{FMM |title=Square Free Sequence |problem=  Prove that no number in the sequence  11, 111, 1111, 11111, ...  is the square of an integer.  |solution=  If s is a number in the...&quot;</title>
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		<updated>2019-11-16T22:31:53Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot;{{FMM |title=Square Free Sequence |problem=  Prove that no number in the sequence  11, 111, 1111, 11111, ...  is the square of an integer.  |solution=  If s is a number in the...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;{{FMM&lt;br /&gt;
|title=Square Free Sequence&lt;br /&gt;
|problem=&lt;br /&gt;
&lt;br /&gt;
Prove that no number in the sequence&lt;br /&gt;
&lt;br /&gt;
11, 111, 1111, 11111, ...&lt;br /&gt;
&lt;br /&gt;
is the square of an integer.&lt;br /&gt;
&lt;br /&gt;
|solution=&lt;br /&gt;
&lt;br /&gt;
If s is a number in the sequence, s must have the form&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;&lt;br /&gt;
11 + 100m&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where m is a non-negative integer. This can be rearranged into two parts: the part divisible by 4, and the part not divisible by 4:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;&lt;br /&gt;
11 + 100m = 100m + 8 + 3 = 4 (25m + 2) + 3&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
which means that when divided by 4, all numbers in the sequence have a remainder of 3.&lt;br /&gt;
&lt;br /&gt;
Furthermore, we know that all squares are of the form&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;&lt;br /&gt;
4n^2&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
or &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;&lt;br /&gt;
4n^2 + 4n + 1&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
and therefore leave remainders of 0 or 1 when divided by 4.&lt;br /&gt;
&lt;br /&gt;
Thus, a number s in the sequence that always has a remainder of 3 cannot be a square.&lt;br /&gt;
&lt;br /&gt;
}}&lt;/div&gt;</summary>
		<author><name>Admin</name></author>
	</entry>
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