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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>.
 +
==Video Solution For Problems 1-3==
 +
https://www.youtube.com/watch?v=5HAk-6qlOH0
  
−
==Solution==
+
== Video Solution by OmegaLearn ==
 +
https://youtu.be/3bRjcrkd5mQ?t=1096
  
−
We call the positive integer mentioned <math>a</math>.  Then <math>a^3 = 16p+1</math>.
+
~ pi_is_3.14
  
−
<math>a^3 = 16p+1</math>
+
==Video Solution by WhyMath==
 +
https://youtu.be/vajttONefxs
  
−
<math>a^3-1 = 16p</math>
+
~savannahsolver
  
−
Factoring the left side:
+
==Solution 1==
  
−
<math>(a-1)\cdot(a^2+a+1) = 16p</math>
+
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.
  
−
We can then try setting one of the factors to <math>16</math>, starting with <math>a-1</math>.
+
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):
  
−
We get <math>a=16+1=17</math>
+
<cmath>\begin{align*}
 +
a^3-1 &= 16p\\
 +
(a-1)(a^2+a+1) &= 16p\\
 +
\end{align*}</cmath>
  
−
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>.
+
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>.
 +
 
 +
Then our other factor, <math>a^2+a+1</math>, is the prime <math>p</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>.
 +
 
 +
Therefore, <math>p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}</math>.
 +
 
 +
==Solution 4==
 +
 
 +
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>.
 +
 
 +
==Solution 5==
 +
 
 +
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>.
 +
 
 +
==Solution 6==
 +
Notice that the cube 16p+1 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, <math>p=\boxed{307}</math> is our only answer
 +
 
 +
pi_is_3.141
 +
 
 +
==Solution 7==
 +
If <math>16p+1 =k</math>, we have <math>k \equiv 1 \mod 16</math>, so <math>k^3 \equiv 1 \mod 16</math>. If <math>k=1</math> we have <math>p=0</math>, which is not prime. If <math>k=17</math> we have <math>16p+1=4913</math>, or <math>p=\boxed{307}</math>
 +
 
 +
==Solution 8 (Pattern Recognition)==
 +
Notice that:
 +
<cmath>\begin{align*}
 +
1^3 &= 0 +1\\
 +
2^3 &= 1*7+1 \\
 +
3^3 &= 2*13+1\\
 +
4^3 &= 3*21+1\\
 +
5^3 &= 4*31+1\\
 +
6^3 &= 5*43+1
 +
\end{align*}</cmath>
 +
 
 +
Here, we can see a clear pattern that <math>n^3=(n-1)p+1</math>, where <math>p</math> is some positive (not necessarily prime) integer. Hence, the equation <math>16p+1=a^3</math> can interpret as <math>17^3 = 16p+1</math>. Solving it, we got <math>p=307</math>. After checking all possible divisors, we will find that <math>307</math> is prime. Hence, we got <math>p=\boxed{307}</math>.
 +
 
 +
==Solution 9 (Slightly Different Modular Arithmetic)==
 +
 
 +
We see that <math>16p+1=n^3</math> for a positive integer <math>n</math>. Subtracting <math>1</math>, we can turn this equation into a modular congruence, since <math>n^3-1</math> must be a multiple of <math>16</math>.
 +
 
 +
Since <math>n^3-1\equiv0\pmod{16}</math>, <math>n^3\equiv1\pmod{16}</math>. We observe that <math>n=1</math> is a solution to this congruence, which doesn't work. The next, or most obvious number to try is <math>n=17</math>. Plugging this in to our original equation, we get
 +
 
 +
<math>17^3=16p+1</math>, yielding <math>p=\boxed{307}</math>, which is prime.
 +
 
 +
 
 +
-among us (countmath1)
 +
 
 +
==Solution 10 (Lengthy Brute Force)==
 +
Recognizing that AIME answers are <math>0</math> through <math>999</math>, the numbers whose cube could even be in contention to be equal to <math>16p + 1</math> are <math>4-25</math>. The cubes of <math>1-3</math> are all below <math>17</math>. We might consider <math>1</math>, but that would result in a <math>p</math> of <math>0</math>, which is not prime and does not follow our given conditions. We can also see that <math>16p + 1</math> must be an odd number. Hence, we brute force by looking at all cubes of odd numbers in <math>5-25</math> until we get that the cube of <math>17</math>, <math>4913</math>, works when <math>p=\boxed{307}</math>
 +
 
 +
Solution by: armang32324
 +
 
 +
== See also ==
 +
{{AIME box|year=2015|n=I|num-b=2|num-a=4}}
 +
{{MAA Notice}}
 +
[[Category:Introductory Number Theory Problems]]

Latest revision as of 22:19, 23 January 2024

Problem

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

Video Solution For Problems 1-3

https://www.youtube.com/watch?v=5HAk-6qlOH0

Video Solution by OmegaLearn

https://youtu.be/3bRjcrkd5mQ?t=1096

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/vajttONefxs

~savannahsolver

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 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

Solution 7

If $16p+1 =k$, we have $k \equiv 1 \mod 16$, so $k^3 \equiv 1 \mod 16$. If $k=1$ we have $p=0$, which is not prime. If $k=17$ we have $16p+1=4913$, or $p=\boxed{307}$

Solution 8 (Pattern Recognition)

Notice that: \begin{align*} 1^3 &= 0 +1\\ 2^3 &= 1*7+1 \\ 3^3 &= 2*13+1\\ 4^3 &= 3*21+1\\ 5^3 &= 4*31+1\\ 6^3 &= 5*43+1 \end{align*}

Here, we can see a clear pattern that $n^3=(n-1)p+1$, where $p$ is some positive (not necessarily prime) integer. Hence, the equation $16p+1=a^3$ can interpret as $17^3 = 16p+1$. Solving it, we got $p=307$. After checking all possible divisors, we will find that $307$ is prime. Hence, we got $p=\boxed{307}$.

Solution 9 (Slightly Different Modular Arithmetic)

We see that $16p+1=n^3$ for a positive integer $n$. Subtracting $1$, we can turn this equation into a modular congruence, since $n^3-1$ must be a multiple of $16$.

Since $n^3-1\equiv0\pmod{16}$, $n^3\equiv1\pmod{16}$. We observe that $n=1$ is a solution to this congruence, which doesn't work. The next, or most obvious number to try is $n=17$. Plugging this in to our original equation, we get

$17^3=16p+1$, yielding $p=\boxed{307}$, which is prime.


-among us (countmath1)

Solution 10 (Lengthy Brute Force)

Recognizing that AIME answers are $0$ through $999$, the numbers whose cube could even be in contention to be equal to $16p + 1$ are $4-25$. The cubes of $1-3$ are all below $17$. We might consider $1$, but that would result in a $p$ of $0$, which is not prime and does not follow our given conditions. We can also see that $16p + 1$ must be an odd number. Hence, we brute force by looking at all cubes of odd numbers in $5-25$ until we get that the cube of $17$, $4913$, works when $p=\boxed{307}$

Solution by: armang32324

See also

2015 AIME I (Problems • Answer Key • Resources)
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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