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

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==Another Another Solution==
 
==Another Another Solution==
Let <math>16p+1=a^3</math>. Realize that <math>a</math> is <math>1(mod 4)</math>. Then we expand, divide both sides by 4, and get <math>16a^3+12a^2+3a=4p</math>. Clearly <math>a=4m</math> for some <math>m</math>. Then, after substitution and another division by 4, we get <math>256m^3+48m^4+3m=p</math>. Since <math>p</math> is prime and there is a factor of <math>m</math> in the LHS, <math>m=1</math>. Therefore, <math>p=\boxed{307}</math>.
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Let <math>16p+1=a^3</math>. Realize that <math>a</math> is <math>1(mod 4)</math>. Expansion, then division by 4, gets <math>16a^3+12a^2+3a=4p</math>. Clearly <math>a=4m</math> for some <math>m</math>. Substitution and another division by 4 gets <math>256m^3+48m^4+3m=p</math>. Since <math>p</math> is prime and there is a factor of <math>m</math> in the LHS, <math>m=1</math>. Therefore, <math>p=\boxed{307}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:53, 7 October 2017

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):

\begin{align*} a^3-1 &= 16p\\ (a-1)(a^2+a+1) &= 16p\\ \end{align*}

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 $p$:

\begin{align*} (a-1)(a^2+a+1) &= 16p\\ (17-1)(17^2+17+1) &=16p\\ p = 289+17+1 &= \boxed{307} \end{align*}

Another Solution

Since $16p+1$ is odd, let $16p+1 = (2a+1)^3$. Therefore, $16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1$. From this, we get $8p=a(4a^2+6a+3)$. We know $p$ is a prime number and it is not an even number. Since $4a^2+6a+3$ is an odd number, we know that $a=8$.

Therefore, $p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}$.

Another Another Solution

Let $16p+1=a^3$. Realize that $a$ is $1(mod 4)$. Expansion, then division by 4, gets $16a^3+12a^2+3a=4p$. Clearly $a=4m$ for some $m$. Substitution and another division by 4 gets $256m^3+48m^4+3m=p$. Since $p$ is prime and there is a factor of $m$ in the LHS, $m=1$. Therefore, $p=\boxed{307}$.

See also

2015 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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