# Difference between revisions of "2015 AIME I Problems/Problem 3"

(Created page with "==Problem== There is a prime number <math>p</math> such that <math>16p+1</math> is the cube of a positive integer. Find <math>p</math>. ==Solution== We call the positive i...") |
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We get <math>a=16+1=17</math> | We get <math>a=16+1=17</math> | ||

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

## Revision as of 11:38, 20 March 2015

## Problem

There is a prime number such that is the cube of a positive integer. Find .

## Solution

We call the positive integer mentioned . Then .

Factoring the left side:

We can then try setting one of the factors to , starting with .

We get

Then our other factor is . A quick divisibility search shows that is prime, so our answer is .