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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

Rewrite as a difference of squares:

n^4 - 360n^2 + 400

= (n^2 + 20)^2 - (20n)^2

Factor it; for the product to be prime, the smaller factor must equal

1

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,

n = 1

n = 19.

Then

f(1) = 41

and

f(19) = 761

are both prime, summing to

802.

Thus, the correct answer is E.

Problem 19 in other years