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2016 AIME I Problems/Problem 12

Revision as of 18:08, 17 February 2020 by Alexlikemath (talk | contribs)

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

$m(m-1)$ 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 $11$.

Let $m^2 - m + 11 = pqrs$ 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 $m \ge 132$. $132^2 - 132 + 11 = 11^3 \cdot 13$, so $m = 132$ and we are done.

Solution 2

Let $m=11k$, then $s=m^2-m+11=11(11k^2-k+1)$. We can see $k = 1 \mod 11$ for $s$ to have a second factor of 11. Let $k=12$, we get $11k^2-k+1=11*11*13$, so $m=11*12=132$. -Mathdummy

Solution 3

First, we can show that $m^2 - m + 11 \not |$ $2,3,5,7$. This can be done by just testing all residue classes.

For example, we can test $m \equiv 0 \mod 2$ or $m \equiv 1 \mod 2$ to show that $m^2 - m + 11$ is not divisible by 2.

Now, we can test $m^2 - m + 11 = 11^4$, which fails, so we test $m^2 - m + 11 = 11^3 \cdot 13$, and we get m = $132$.

-AlexLikeMath

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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