https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_83&feed=atom&action=history2008 iTest Problems/Problem 83 - Revision history2024-03-28T21:08:06ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_83&diff=98939&oldid=prevRockmanex3 at 01:23, 23 November 20182018-11-23T01:23:52Z<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 01:23, 23 November 2018</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>&= \frac{n(2n+1)(7n+1)}{6}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>&= \frac{n(2n+1)(7n+1)}{6}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\end{align*}</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\end{align*}</cmath></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Thus, <math>(1^<del class="diffchange diffchange-inline">2+2^2+3</del>^2<del class="diffchange diffchange-inline">+</del>\<del class="diffchange diffchange-inline">cdots + n^2</del>)\left<del class="diffchange diffchange-inline">[</del>(n+1<del class="diffchange diffchange-inline">)^2+(n+2)^2+(n+3)</del>^<del class="diffchange diffchange-inline">2+\cdots + (</del>2n<del class="diffchange diffchange-inline">)</del>^2\right<del class="diffchange diffchange-inline">] </del>= \frac{n^2 (2n+1)^2 (n+1)(7n+1)}{36}</math>. In order for the expression to be a perfect square, <math>(n+1)(7n+1)</math> must be a perfect square.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Thus, <math><ins class="diffchange diffchange-inline">\left</ins>( <ins class="diffchange diffchange-inline">\sum_{i=</ins>1<ins class="diffchange diffchange-inline">}</ins>^<ins class="diffchange diffchange-inline">n i</ins>^2 \<ins class="diffchange diffchange-inline">right</ins>)\left(<ins class="diffchange diffchange-inline">\sum_{i=</ins>n+1<ins class="diffchange diffchange-inline">}</ins>^<ins class="diffchange diffchange-inline">{</ins>2n<ins class="diffchange diffchange-inline">} i</ins>^2 \right<ins class="diffchange diffchange-inline">) </ins>= \frac{n^2 (2n+1)^2 (n+1)(7n+1)}{36}</math>. In order for the expression to be a perfect square, <math>(n+1)(7n+1)</math> must be a perfect square.</div></td></tr>
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</table>Rockmanex3https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_83&diff=98906&oldid=prevRockmanex3: Solution to Problem 83 -- squares, number theory, and casework2018-11-22T01:23:35Z<p>Solution to Problem 83 -- squares, number theory, and casework</p>
<p><b>New page</b></p><div>==Problem==<br />
<br />
Find the greatest natural number <math>n</math> such that <math>n\leq 2008</math> and <math>(1^2+2^2+3^2+\cdots + n^2)\left[(n+1)^2+(n+2)^2+(n+3)^2+\cdots + (2n)^2\right]</math> is a perfect square.<br />
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==Solution==<br />
<br />
Notice that <math>\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}</math>, so<br />
<cmath>\begin{align*}<br />
\sum_{i=n+1}^{2n} i^2 &= \sum_{i=1}^{2n} i^2 - \sum_{i=1}^n i^2 \\<br />
&= \frac{2n(2n+1)(4n+1)}{6} - \frac{n(n+1)(2n+1)}{6} \\<br />
&= \frac{16n^3 + 12n^2 + 2n}{6} - \frac{2n^3 + 3n^2 + n}{6} \\<br />
&= \frac{14n^3 + 9n^2 + n}{6} \\<br />
&= \frac{n(2n+1)(7n+1)}{6}<br />
\end{align*}</cmath><br />
Thus, <math>(1^2+2^2+3^2+\cdots + n^2)\left[(n+1)^2+(n+2)^2+(n+3)^2+\cdots + (2n)^2\right] = \frac{n^2 (2n+1)^2 (n+1)(7n+1)}{36}</math>. In order for the expression to be a perfect square, <math>(n+1)(7n+1)</math> must be a perfect square.<br />
<br />
<br><br />
By using the [[Euclidean algorithm|Euclidean Algorithm]], <math>\gcd(n+1,7n+1) = \gcd(n+1,6)</math>. Thus, the [[GCD]] of <math>n+1</math> and <math>7n+1</math> must be factors of 6. Now, split the factors as different [[casework]]. Note that the [[quadratic residues]] of 7 are 0, 1, 2, and 4.<br />
<br />
* If <math>\gcd(n+1,7n+1) = 6</math>, then <math>n \equiv 5 \pmod{6}</math>. Let <math>n = 6a+5</math>, so <math>(n+1)(7n+1) = (6a+6)(42a+36) = 36(a+1)(7a+6)</math>. Since 6 is divided out of <math>n+1</math> and <math>7n+1</math>, <math>a+1</math> and <math>7a+6</math> are relatively prime, so <math>a+1</math> and <math>7a+6</math> must be perfect squares. However, since 6 is not a quadratic residue of 7, the GCD of <math>n+1</math> and <math>7n+1</math> can not be 6.<br />
* If <math>\gcd(n+1,7n+1) = 3</math>, then <math>n \equiv 2 \pmod{3}</math>. Let <math>n = 3a+2</math>, so <math>(n+1)(7n+1) = (3a+3)(21a+15) = 9(a+1)(7a+5)</math>. Since 3 is divided out of <math>n+1</math> and <math>7n+1</math>, <math>a+1</math> and <math>7a+5</math> are relatively prime, so <math>a+1</math> and <math>7a+5</math> must be perfect squares. However, since 5 is not a quadratic residue of 7, the GCD of <math>n+1</math> and <math>7n+1</math> can not be 3.<br />
* If <math>\gcd(n+1,7n+1) = 2</math>, then <math>n \equiv 1 \pmod{2}</math>. Let <math>n = 2a+1</math>, so <math>(n+1)(7n+1) = (2a+2)(14a+8) = 4(a+1)(7a+4)</math>. Since 2 is divided out of <math>n+1</math> and <math>7n+1</math>, <math>a+1</math> and <math>7a+4</math> are relatively prime, so <math>a+1</math> and <math>7a+4</math> must be perfect squares. We also know that <math>n+1</math> and <math>7n+1</math> do not share a factor of 3, so <math>n \equiv 1,3 \pmod{6}</math>. That means <math>n \le 2007</math>, so <math>a \le 1003</math>. After trying values of <math>a</math> that are one less than a perfect square, we find that the largest value that makes <math>(n+1)(7n+1)</math> a perfect square is <math>a = 960</math>. That means <math>n = 1921</math>.<br />
* If <math>\gcd(n+1,7n+1) = 1</math>, then <math>n+1 \equiv 1,5 \pmod{6}</math> (to avoid common factors that are factors of 6), so <math>n \equiv 0,4 \pmod{6}</math>. After trying values of <math>n</math> that are one less than a perfect square, we find that the largest value that makes <math>(n+1)(7n+1)</math> a perfect square is <math>n = 120</math> (we could also stop searching once <math>n</math> gets below 1921).<br />
<br />
From the casework, the largest natural number <math>n</math> that makes <math>(1^2+2^2+3^2+\cdots + n^2)\left[(n+1)^2+(n+2)^2+(n+3)^2+\cdots + (2n)^2\right]</math> is a perfect square is <math>\boxed{1921}</math>.<br />
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==See Also==<br />
{{2008 iTest box|num-b=82|num-a=84}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
[[Category:Intermediate Number Theory Problems]]</div>Rockmanex3