2009 AMC 12B Problem 19

Attempt Problem 19 of the 2009 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

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19.

For each positive integer $n,$ let $f(n) = n^4 - 360n^2 + 400.$ What is the sum of all values of $f(n)$ that are prime numbers?

$794$

$796$

$798$

$800$

$802$

Answer: E

Concepts:difference of squares factoring prime

Difficulty rating: 2000

Solution:

Write $\begin{aligned} f(n) &= n^4 + 40n^2 + 400 - 400n^2 \\ &= (n^2 + 20)^2 - (20n)^2 \\ &= (n^2 + 20n + 20) \\ &\quad {}\cdot (n^2 - 20n + 20). \end{aligned}$

For $f(n)$ to be prime the smaller factor must be $1$ : solving $n^2 - 20n + 20 = 1$ gives $(n - 1)(n - 19) = 0,$ so $n = 1$ or $n = 19.$

Then $f(1) = 41$ and $f(19) = 761$ are both prime, summing to $802.$

Thus, the correct answer is E.

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