2015 AIME I Problems/Problem 3

Revision as of 12:38, 20 March 2015 by Pieater314159 (talk | contribs) (Solution)


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


We call the positive integer mentioned $a$. Then $a^3 = 16p+1$.

$a^3 = 16p+1$

$a^3-1 = 16p$

Factoring the left side:

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

We can then try setting one of the factors to $16$, starting with $a-1$.

We get $a=16+1=17$

Then our other factor is $a^2+a+1=17^2+17+1=289+17+1=307$. A quick divisibility search shows that $307$ is prime, so our answer is $\boxed{307}$.

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