Problem 26 · 2019 Math Kangaroo
Stretch
Number Theory
factorizationprime-test
For how many integers \(n\) is \(|n^{2} - 2n - 3|\) a prime number?
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Answer: D — 4
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Hint 1 of 2
Factor \(n^{2} - 2n - 3 = (n-3)(n+1)\); a product of integers is prime only if one factor is \(\pm 1\).
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Hint 2 of 2
Set each factor to \(\pm 1\) and check whether the other factor's absolute value is prime.
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Approach: factor, then force one factor to be \(\pm 1\)
- \(|n^{2} - 2n - 3| = |(n-3)(n+1)|\); for this to be prime, one factor must be \(\pm 1\).
- \(n - 3 = \pm 1\) gives \(n = 4\) (value 5) or \(n = 2\) (value 3); \(n + 1 = \pm 1\) gives \(n = 0\) (value 3) or \(n = -2\) (value 5).
- All four results (3 or 5) are prime, so there are 4 such integers: \(n = -2, 0, 2, 4\).
- Answer (D) 4.
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