2016 AIME I Problems/Problem 12

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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 $m = \fbox{132}$.

See Also

2016 AIME I (ProblemsAnswer KeyResources)
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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