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

Revision as of 13:23, 26 July 2021

Problem

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

Video Solution

~ pi_is_3.14

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

Solution 7

We make a table of $x^3$ (mod 16). We notice that only $x\cong1$ (mod 16) gives $x^3\cong1$ (mod 16). Then, we try $16p+1=1$ , which obviously doesn't work. However, $16p+1=4913$ gives $p=\fbox{307}$ .

~john0512

@@ Line 83: / Line 83: @@
 pi_is_3.141
+==Solution 7==
+We make a table of <math>x^3</math> (mod 16). We notice that only <math>x\cong1</math> (mod 16) gives <math>x^3\cong1</math> (mod 16). Then, we try <math>16p+1=1</math>, which obviously doesn't work. However, <math>16p+1=4913</math> gives <math>p=\fbox{307}</math>.
+~john0512
 == See also ==

2015 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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