2015 AIME I Problems/Problem 3

Problem

There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$ .

Solution

Let the positive integer mentioned be $a$ , so that $a^3 = 16p+1$ . Note that $a$ must be odd, because $16p+1$ is odd.

Rearrange this expression and factor the left side (this factoring can be done using $(a^3-b^3) = (a-b)(a^2+a b+b^2)$ , or synthetic divison once it is realized that $a = 1$ is a root):

$a^3-1 = 16p$

$(a-1)(a^2+a+1) = 16p$

Because $a$ is odd, $a-1$ is even and $a^2+a+1$ is odd. If $a^2+a+1$ is odd, $a-1$ must be some multiple of $16$ . However, for $a-1$ to be any multiple of $16$ other than $16$ would mean $p$ is not a prime. Therefore, $a-1 = 16$ and $a = 17$ .

Then our other factor, $a^2+a+1$ , is the prime $p$ :

$(a-1)(a^2+a+1) = 16p$

$(17-1)(17^2+17+1) =16p$

$p = 289+17+1 = \boxed{307}$ .

2015 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2015 AIME I Problems/Problem 3

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