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

Revision as of 12:48, 30 May 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 obviously 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

@@ Line 24: / Line 24: @@
 \end{align*}</cmath>
-==Solution 2==
+==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*}
+p &= n^3-1\\
+p &= n^3-1^3\\
+p &= (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\\
+&= n^2+n+1\\
+n^2+n &= 15\\
+n(n+1) &= 15\\
+\end{align*}</cmath>
+Which is obviously impossible for integer n. So <math>p=n^2+n+1</math> and
+<cmath>\begin{align*}
+&= 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>.
@@ Line 30: / Line 56: @@
 Therefore, <math>p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}</math>.
-==Solution 3==
+==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 4==
+==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 5==
+==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

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