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Difference between revisions of "2015 AIME I Problems/Problem 3"
m (sol more readable)
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Let the positive integer mentioned be <math>a</math>, so that <math>a^3 = 16p+1</math>. Note that <math>a</math> must be odd, because <math>16p+1</math> is odd.
 
Let the positive integer mentioned be <math>a</math>, so that <math>a^3 = 16p+1</math>. Note that <math>a</math> must be odd, because <math>16p+1</math> is odd.
  
−
Rearrange this expression and factor the left side (this factoring can be done using <math>(a^3-b^3) = (a-b)(a^2+a b+b^2)</math>, or synthetic divison once it is realized that <math>a = 1</math> is a root):
+
Rearrange this expression and factor the left side (this factoring can be done using <math>(a^3-b^3) = (a-b)(a^2+a b+b^2)</math> or synthetic divison once it is realized that <math>a = 1</math> is a root):
  
−
<math>a^3-1 = 16p</math>
+
<cmath>\begin{align*}
−
 
+
a^3-1 &= 16p\\
−
<math>(a-1)(a^2+a+1) = 16p</math>
+
(a-1)(a^2+a+1) &= 16p\\
 +
\end{align*}</cmath>
  
 
Because <math>a</math> is odd, <math>a-1</math> is even and <math>a^2+a+1</math> is odd. If <math>a^2+a+1</math> is odd, <math>a-1</math> must be some multiple of <math>16</math>. However, for <math>a-1</math> to be any multiple of <math>16</math> other than <math>16</math> would mean <math>p</math> is not a prime. Therefore, <math>a-1 = 16</math> and <math>a = 17</math>.
 
Because <math>a</math> is odd, <math>a-1</math> is even and <math>a^2+a+1</math> is odd. If <math>a^2+a+1</math> is odd, <math>a-1</math> must be some multiple of <math>16</math>. However, for <math>a-1</math> to be any multiple of <math>16</math> other than <math>16</math> would mean <math>p</math> is not a prime. Therefore, <math>a-1 = 16</math> and <math>a = 17</math>.
Line 17: Line 18:
 
Then our other factor, <math>a^2+a+1</math>, is the prime <math>p</math>:
 
Then our other factor, <math>a^2+a+1</math>, is the prime <math>p</math>:
  
−
<math>(a-1)(a^2+a+1) = 16p</math>
+
<cmath>\begin{align*}
−
 
+
(a-1)(a^2+a+1) &= 16p\\
−
<math>(17-1)(17^2+17+1) =16p</math>
+
(17-1)(17^2+17+1) &=16p\\
−
 
+
p = 289+17+1 &= \boxed{307}
−
<math>p = 289+17+1 = \boxed{307}</math>.
+
\end{align*}</cmath>
  
 
==Another Solution==
 
==Another Solution==
  
−
Since <math>16p+1</math> is odd, let <math>16p+1 = (2a+1)^3</math>  
+
Since <math>16p+1</math> is odd, let <math>16p+1 = (2a+1)^3</math>. Therefore, <math>16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1</math>. From this, we get <math>8p=a(4a^2+6a+3)</math>. We know <math>p</math> is a prime number and it is not an even number. Since <math>4a^2+6a+3</math> is an odd number, we know that <math>a=8</math>.
−
 
 
−
<math>16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1</math>
 
−
 
 
−
We get:
 
−
 
 
−
<math>8p=a(4a^2+6a+3)</math>
 
−
 
 
−
We know p is a prime number and apparently not an even number.
 
−
and <math>4a^2+6a+3</math> is an odd number, so a must equal 8.
 
  
−
so we get <math>p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}</math>.
+
Therefore, <math>p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:27, 21 February 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, $p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}$.

See also

2015 AIME I (Problems • Answer Key • Resources)
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Problem 4
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