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

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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>.
 
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==
+
==Solution 1==
  
 
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>.
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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\\
 +
(17-1)(17^2+17+1) &=16p\\
 +
p = 289+17+1 &= \boxed{307}
 +
\end{align*}</cmath>
 +
 
 +
==Solution 2 (Similar to 1)==
 +
 
 +
Observe that this is the same as <math>16p+1=n^3</math> for some integer <math>n</math>.
 +
So:
 +
 
 +
<cmath>\begin{align*}
 +
16p &= n^3-1\\
 +
16p &= n^3-1^3\\
 +
16p &= (n-1)(n^2+n+1)\\
 +
\end{align*}</cmath>
 +
 
 +
Observe that either <math>p=n-1</math> or <math>p=n^2+n+1</math> because <math>p</math> and <math>16</math> share no factors (<math>p</math> can't be <math>2</math>).
 +
Let <math>p=n-1</math>.
 +
Then:
 +
 
 +
<cmath>\begin{align*}
 +
p &= n-1\\
 +
16 &= n^2+n+1\\
 +
n^2+n &= 15\\
 +
n(n+1) &= 15\\
 +
\end{align*}</cmath>
 +
 
 +
Which is impossible for integer n. So <math>p=n^2+n+1</math> and
 +
 
 +
<cmath>\begin{align*}
 +
16 &= n-1\\
 +
n &= 17\\
 +
p &= 17^2+17+1\\
 +
p = 289+17+1 &= \boxed{307}\\
 +
\end{align*}</cmath> - firebolt360
 +
 
 +
==Solution 3==
 +
 
 +
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>(17-1)(17^2+17+1) =16p</math>
+
Therefore, <math>p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}</math>.
  
<math>p = 289+17+1 = \boxed{307}</math>.
+
==Solution 4==
  
==Another Solution==
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Let <math>16p+1=a^3</math>. Realize that <math>a</math> congruent to <math>1\mod 4</math>, so let <math>a=4n+1</math>. Expansion, then division by 4, gets <math>16n^3+12n^2+3n=4p</math>. Clearly <math>n=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>.
  
Since <math>16p+1</math> is odd, let <math>16p+1 = (2a+1)^3</math>
+
==Solution 5==
  
<math>16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1</math>
+
Notice that <math>16p+1</math> must be in the form <math>(a+1)^3 = a^3 + 3a^2 + 3a + 1</math>. Thus <math>16p = a^3 + 3a^2 + 3a</math>, or <math>16p = a\cdot (a^2 + 3a + 3)</math>. Since <math>p</math> must be prime, we either have <math>p = a</math> or <math>a = 16</math>. Upon further inspection and/or using the quadratic formula, we can deduce <math>p \neq a</math>. Thus we have <math>a = 16</math>, and <math>p = 16^2 + 3\cdot 16 + 3 = \boxed{307}</math>.
  
We get:
+
==Solution 6==
 +
Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15.
 +
Case one: The cube is of the form 16k+1-->Plugging this in, and taking note that p is prime and has only 1 factor gives p=307
 +
Case two: The cube is of the form 16k+15--> Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here
  
<math>8p=a(4a^2+6a+3)</math>
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Hence, <math>p=\boxed{307}</math> is our only answer
  
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>.
+
pi_is_3.141
  
 
== See also ==
 
== See also ==

Latest revision as of 04:39, 1 June 2020

Problem

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

Solution 1

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*}

Solution 2 (Similar to 1)

Observe that this is the same as $16p+1=n^3$ for some integer $n$. So:

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

Observe that either $p=n-1$ or $p=n^2+n+1$ because $p$ and $16$ share no factors ($p$ can't be $2$). Let $p=n-1$. Then:

\begin{align*} p &= n-1\\ 16 &= n^2+n+1\\ n^2+n &= 15\\ n(n+1) &= 15\\ \end{align*}

Which is impossible for integer n. So $p=n^2+n+1$ and

\begin{align*} 16 &= n-1\\ n &= 17\\ p &= 17^2+17+1\\ p = 289+17+1 &= \boxed{307}\\ \end{align*} - firebolt360

Solution 3

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

Solution 4

Let $16p+1=a^3$. Realize that $a$ congruent to $1\mod 4$, so let $a=4n+1$. Expansion, then division by 4, gets $16n^3+12n^2+3n=4p$. Clearly $n=4m$ 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, $p=\boxed{307}$.

Solution 5

Notice that $16p+1$ must be in the form $(a+1)^3 = a^3 + 3a^2 + 3a + 1$. Thus $16p = a^3 + 3a^2 + 3a$, or $16p = a\cdot (a^2 + 3a + 3)$. Since $p$ must be prime, we either have $p = a$ or $a = 16$. Upon further inspection and/or using the quadratic formula, we can deduce $p \neq a$. Thus we have $a = 16$, and $p = 16^2 + 3\cdot 16 + 3 = \boxed{307}$.

Solution 6

Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. Case one: The cube is of the form 16k+1-->Plugging this in, and taking note that p is prime and has only 1 factor gives p=307 Case two: The cube is of the form 16k+15--> Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here

Hence, $p=\boxed{307}$ is our only answer


pi_is_3.141

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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