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Difference between revisions of "2016 AIME I Problems/Problem 12"

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We claim <math>m = 132</math>. Note <math>132^2 - 132 + 11 = 11^3 \cdot 13</math>.
 
We claim <math>m = 132</math>. Note <math>132^2 - 132 + 11 = 11^3 \cdot 13</math>.
  
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Suppose <math>m < 132</math> and <math>m^2 - m + 11 = pqrs</math> for primes <math>p, q, r, s</math>. We easily verify that <math>p, q, r, s \ge 11</math> using quadratic residue argument. But <math>p, q, r, s = 11</math> yields no integer solution for <math>m</math>. Thus <math>pqrs \ge 11^3 \cdot 13</math>. But this requires <math>m \ge 132</math> from solving the quadratic, contradiction. Hence <math>m = \fbox{132}</math>.
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Now <math>m(m-1)</math> is the product of two consecutive integers, so it is always even. Thus <math>m(m-1)+11</math> is odd and never divisible by <math>2</math>. Thus any prime <math>p</math> that divides <math>m^2-m+11</math> must divide <math>4m^2-4m+44=(2m-1)^2+43</math>. We see that <math>(2m-1)^2\equiv -43\pmod{p}</math>. We can verify that <math>-43</math> is not a perfect square mod <math>p</math> for each of <math>p=3,5,7</math>. Therefore, all prime factors of <math>m^2-m+11</math> are greater than or equal to <math>11</math>.
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Now suppose <math>m < 132</math> and <math>m^2 - m + 11 = pqrs</math> for primes <math>p, q, r, s\ge11</math>. If <math>p, q, r, s = 11</math>, then <math>m^2-m+11=11^4</math>. We can multiply this by <math>4</math> and complete the square to find <math>(2m-1)^2=4\cdot 11^4-43</math>. But
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<cmath>(2\cdot 11^2-1)^2=4\cdot 11^4-4\cdot 11^2+1 <4\cdot 11^4-43<(2\cdot 11^2)^2,</cmath>
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hence we have pinned a perfect square <math>(2m-1)^2=4\cdot 11^4-43</math> strictly between two consecutive perfect squares, a contradiction. Hence <math>pqrs \ge 11^3 \cdot 13</math>. Thus <math>m^2-m+11\ge 11^3\cdot 13</math>, or <math>(m-132)(m+131)\ge0</math>. From the inequality, we see that <math>m \ge 132</math>, hence <math>m = \fbox{132}</math>.
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==See Also==
 
==See Also==
 
{{AIME box|year=2016|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2016|n=I|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:35, 7 March 2016

Problem

Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes.

Solution

We claim $m = 132$. Note $132^2 - 132 + 11 = 11^3 \cdot 13$.

Now $m(m-1)$ is the product of two consecutive integers, so it is always even. Thus $m(m-1)+11$ is odd and never divisible by $2$. Thus any prime $p$ that divides $m^2-m+11$ must divide $4m^2-4m+44=(2m-1)^2+43$. We see that $(2m-1)^2\equiv -43\pmod{p}$. We can verify that $-43$ is not a perfect square mod $p$ for each of $p=3,5,7$. Therefore, all prime factors of $m^2-m+11$ are greater than or equal to $11$.

Now suppose $m < 132$ and $m^2 - m + 11 = pqrs$ for primes $p, q, r, s\ge11$. If $p, q, r, s = 11$, then $m^2-m+11=11^4$. We can multiply this by $4$ and complete the square to find $(2m-1)^2=4\cdot 11^4-43$. But \[(2\cdot 11^2-1)^2=4\cdot 11^4-4\cdot 11^2+1 <4\cdot 11^4-43<(2\cdot 11^2)^2,\] hence we have pinned a perfect square $(2m-1)^2=4\cdot 11^4-43$ strictly between two consecutive perfect squares, a contradiction. Hence $pqrs \ge 11^3 \cdot 13$. Thus $m^2-m+11\ge 11^3\cdot 13$, or $(m-132)(m+131)\ge0$. From the inequality, we see that $m \ge 132$, hence $m = \fbox{132}$.

See Also

2016 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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