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

Revision as of 23:47, 4 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 $132^2 - 132 + 11 = 11^3 \cdot 13$ .

Suppose $m < 132$ and $m^2 - m + 11 = pqrs$ for primes $p, q, r, s$ . We easily verify that $p, q, r, s \ge 11$ using quadratic residue argument. But $p, q, r, s = 11$ yields no integer solution for $m$ . Thus $pqrs \ge 11^3 \cdot 13$ . But this requires $m \ge 132$ from solving the quadratic, contradiction. Hence $m = \fbox{132}$ .

@@ Line 4: / Line 4: @@
 We claim <math>m = 132</math>. Note <math>132^2 - 132 + 11 = 11^3 \cdot 13</math>.
-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 = 132</math>.
+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>.
 ==See Also==
 {{AIME box|year=2016|n=I|num-b=11|num-a=13}}
 {{MAA Notice}}

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

Revision as of 23:47, 4 March 2016

Problem

Solution

See Also