https://artofproblemsolving.com/wiki/index.php?title=2011_JBMO_Problems/Problem_2&feed=atom&action=history2011 JBMO Problems/Problem 2 - Revision history2024-03-29T09:01:03ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_JBMO_Problems/Problem_2&diff=105097&oldid=prevRockmanex3: Solution to Problem 2 -- finding p2019-04-02T23:56:19Z<p>Solution to Problem 2 -- finding p</p>
<p><b>New page</b></p><div>==Problem==<br />
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Find all primes <math>p</math> such that there exist positive integers <math>x,y</math> that satisfy <math>x(y^2-p)+y(x^2-p)=5p</math>.<br />
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==Solution==<br />
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Rearrange the original equation to get<br />
<cmath>\begin{align*}<br />
5p &= xy^2 + x^2 y - xp - yp \\<br />
&= xy(x+y) -p(x+y) \\<br />
&= (xy-p)(x+y)<br />
\end{align*}</cmath><br />
Since <math>5p</math>, <math>xy-p</math>, and <math>x+y</math> are all integers, <math>xy-p</math> and <math>x+y</math> must be a [[factor]] of <math>5p</math>. Now there are four cases to consider.<br />
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'''Case 1: <math>xy-p = 5p</math> and <math>x+y = 1</math>'''<br />
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Since <math>x</math> and <math>y</math> are positive integers, <math>x+y \ne 1</math>, so the first case can not happen.<br />
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'''Case 2: <math>xy-p = 1</math> and <math>x+y = 5p</math>'''<br />
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Rearranging the first equation results in <math>5xy-5 = 5p</math>. By substitution, we get<br />
<cmath>\begin{align*}<br />
x+y &= 5xy - 5 \\<br />
5 &= 5xy - x - y \\<br />
25 &= 25xy - 5x - 5y.<br />
\end{align*}</cmath><br />
Applying [[Simon's Favorite Factoring Trick]] results in<br />
<cmath>26 = (5x-1)(5y-1)</cmath>.<br />
From this equation, <math>5x-1</math> and <math>5y-1</math> can equal 1, 2, 13, or 26. Since no value of <math>x</math> or <math>y</math> work, there are no solutions, so the second case can not happen.<br />
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'''Case 3: <math>xy-p = 5</math> and <math>x+y = p</math>'''<br />
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Rearranging the first equation results in <math>xy-5 = p</math>. By substitution and Simon's Favorite Factoring Trick, we get<br />
<cmath>\begin{align*}<br />
xy-5 &= x+y \\<br />
xy-x-y &= 5 \\<br />
(x-1)(y-1) &= 6<br />
\end{align*}</cmath><br />
From this equation, <math>x-1</math> and <math>y-1</math> can equal 1, 2, 3, or 6. The ordered pairs of <math>(x,y)</math> can be <math>(2,7)</math>, <math>(7,2)</math>, <math>(4,3)</math>, or <math>(3,4)</math>. Since <math>p</math> is prime, <math>p</math> can equal <math>7</math> in this case.<br />
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'''Case 4: <math>xy-p = p</math> and <math>x+y = 5</math>'''<br />
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Rearranging the first equation results in <math>xy = 2p</math>. Since <math>x,y</math> are positive integers, we can simply take the few ordered pairs <math>(x,y)</math> that satisfy the second equation and plug the values into the first equation to check for values of <math>p</math>.<br />
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If <math>(x,y)</math> is <math>(1,4)</math> or <math>(4,1)</math>, then <math>2p = 4</math>, so <math>p = 2</math>. If <math>(x,y)</math> is <math>(2,3)</math> or <math>(3,2)</math>, then <math>2p = 6</math>, so <math>p = 3</math>. Both <math>2</math> and <math>3</math> are prime numbers, so both of the numbers work for <math>p</math>.<br />
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In summary, the values of <math>p</math> that satisfy the original conditions are <math>\boxed{2, 3, 7}</math>.<br />
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==See Also==<br />
{{JBMO box|year=2011|num-b=1|num-a=3|five=}}<br />
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[[Category:Intermediate Number Theory Problems]]</div>Rockmanex3