2009 AMC 12B Problems/Problem 19

Revision as of 01:00, 11 January 2010 by Not trig (talk | contribs) (Solution 3)

Problem

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?

$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$

Solution

Solution 1

To find the answer it was enough to play around with $f$. One can easily find that $f(1)=41$ is a prime, then $f$ becomes negative for $n$ between $2$ and $18$, and then $f(19)=761$ is again a prime number. And as $41 + 761 = 802$ is already the largest option, the answer must be $\boxed{802}$.

Solution 2

We will now show a complete solution, with a proof that no other values are prime.

Consider the function $g(x) = x^2 - 360x + 400$, then obviously $f(x) = g(x^2)$.

The roots of $g$ are: \[x_{1,2}  = \frac{ 360 \pm \sqrt{ 360^2 - 4\cdot 400 } }2  = 180 \pm 80 \sqrt 5\]

We can then write $g(x) = (x - 180 - 80\sqrt 5)(x - 180 - 80\sqrt 5)$, and thus $f(x) = (x^2 - 180 - 80\sqrt 5)(x^2 - 180 - 80\sqrt 5)$.

We would now like to factor the right hand side further, using the formula $(x^2 - y^2) = (x-y)(x+y)$. To do this, we need to express both constants as squares of some other constants. Luckily, we have a pretty good idea how they look like.

We are looking for rational $a$ and $b$ such that $(a+b\sqrt 5)^2 = 180 + 80\sqrt 5$. Expanding the left hand side and comparing coefficients, we get $ab=40$ and $a^2 + 5b^2 = 180$. We can easily guess (or compute) the solution $a=10$, $b=4$.

Hence $180 + 80\sqrt 5 = (10 + 4\sqrt 5)^2$, and we can easily verify that also $180 - 80\sqrt 5 = (10 - 4\sqrt 5)^2$.

We now know the complete factorization of $f(x)$:

\[f(x) = (x - 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5)(x - 10 + 4\sqrt 5)(x + 10 - 4\sqrt 5)\]

As the final step, we can now combine the factors in a different way, in order to get rid of the square roots.

We have $(x - 10 - 4\sqrt 5)(x - 10 + 4\sqrt 5) = (x-10)^2 - (4\sqrt 5)^2 = x^2 - 20x + 20$, and $(x + 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5) = x^2 + 20x + 20$.

Hence we obtain the factorization $f(x) = (x^2 - 20x + 20)(x^2 + 20x + 20)$.

For $x\geq 20$ both terms are positive and larger than one, hence $f(x)$ is not prime. For $1<x<19$ the second factor is positive and the first one is negative, hence $f(x)$ is not a prime. The remaining cases are $x=1$ and $x=19$. In both cases, $f(x)$ is indeed a prime, and their sum is $f(1) + f(19) = 41 + 761 = \boxed{802}$.

Solution 3

Instead of doing the hard work, we can try to guess the factorization. One good approach:

We can make the observation that $f(x)$ looks similar to $(x^2 + 20)^2$ with the exception of the $x^2$ term. In fact, we have $(x^2 + 20)^2 = x^4 + 40x^2 + 400$. But then we notice that it differs from the desired expression by a square: $f(x) = (x^2 + 20)^2 - 400x^2 = (x^2 + 20)^2 - (20x)^2$.

Now we can use the formula $(x^2 - y^2) = (x-y)(x+y)$ to obtain the same factorization as in the previous solution, without all the work.

See Also

2009 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 12 Problems and Solutions
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