Difference between revisions of "2024 AIME I Problems/Problem 13"

Latest revision as of 17:39, 24 November 2024

Problem

Let $p$ be the least prime number for which there exists a positive integer $n$ such that $n^{4}+1$ is divisible by $p^{2}$ . Find the least positive integer $m$ such that $m^{4}+1$ is divisible by $p^{2}$ .

Solution 1

If $p=2$, then $4\mid n^4+1$ for some integer $n$. But $\left(n^2\right)^2\equiv0$ or $1\pmod4$, so it is impossible. Thus $p$ is an odd prime.

For integer $n$ such that $p^2\mid n^4+1$, we have $p\mid n^4+1$, hence $p\nmid n^4-1$, but $p\mid n^8-1$. By Fermat's Little Theorem, $p\mid n^{p-1}-1$, so \begin{equation*} p\mid\gcd\left(n^{p-1}-1,n^8-1\right)=n^{\gcd(p-1,8)}-1. \end{equation*} Here, $\gcd(p-1,8)$ mustn't be divide into $4$ or otherwise $p\mid n^{\gcd(p-1,8)}-1\mid n^4-1$, which contradicts. So $\gcd(p-1,8)=8$, and so $8\mid p-1$. The smallest such prime is clearly $p=17=2\times8+1$. So we have to find the smallest positive integer $m$ such that $17\mid m^4+1$. We first find the remainder of $m$ divided by $17$ by doing \begin{array}{|c|cccccccccccccccc|} \hline \vphantom{\tfrac11}x\bmod{17}&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\\hline \vphantom{\dfrac11}\left(x^4\right)+1\bmod{17}&2&0&14&2&14&5&5&0&0&5&5&14&2&14&0&2\\\hline \end{array} So $m\equiv\pm2$, $\pm8\pmod{17}$. If $m\equiv2\pmod{17}$, let $m=17k+2$, by the binomial theorem, \begin{align*} 0&\equiv(17k+2)^4+1\equiv\mathrm {4\choose 1}(17k)(2)^3+2^4+1=17(1+32k)\pmod{17^2}\\[3pt] \implies0&\equiv1+32k\equiv1-2k\pmod{17}. \end{align*} So the smallest possible $k=9$, and $m=155$.

If $m\equiv-2\pmod{17}$, let $m=17k-2$, by the binomial theorem, \begin{align*} 0&\equiv(17k-2)^4+1\equiv\mathrm {4\choose 1}(17k)(-2)^3+2^4+1=17(1-32k)\pmod{17^2}\\[3pt] \implies0&\equiv1-32k\equiv1+2k\pmod{17}. \end{align*} So the smallest possible $k=8$, and $m=134$.

If $m\equiv8\pmod{17}$, let $m=17k+8$, by the binomial theorem, \begin{align*} 0&\equiv(17k+8)^4+1\equiv\mathrm {4\choose 1}(17k)(8)^3+8^4+1=17(241+2048k)\pmod{17^2}\\[3pt] \implies0&\equiv241+2048k\equiv3+8k\pmod{17}. \end{align*} So the smallest possible $k=6$, and $m=110$.

If $m\equiv-8\pmod{17}$, let $m=17k-8$, by the binomial theorem, \begin{align*} 0&\equiv(17k-8)^4+1\equiv\mathrm {4\choose 1}(17k)(-8)^3+8^4+1=17(241-2048k)\pmod{17^2}\\[3pt] \implies0&\equiv241+2048k\equiv3+9k\pmod{17}. \end{align*} So the smallest possible $k=11$, and $m=179$.

In conclusion, the smallest possible $m$ is $\boxed{110}$.

Solution by Quantum-Phantom

Solution 2

We work in the ring $\mathbb Z/289\mathbb Z$ and use the formula $\sqrt[4]{-1}=\pm\sqrt{\frac12}\pm\sqrt{-\frac12}.$ Since $-\frac12=144$, the expression becomes $\pm12\pm12i$, and it is easily calculated via Hensel that $i=38$, thus giving an answer of $\boxed{110}$.

Solution 3 (Easy, given specialized knowledge)

Note that $n^4 + 1 \equiv 0 \pmod{p}$ means $\text{ord}_{p}(n) = 8 \mid p-1.$ The smallest prime that does this is $17$ and $2^4 + 1 = 17$ for example. Now let $g$ be a primitive root of $17^2.$ The satisfying $n$ are of the form, $g^{\frac{p(p-1)}{8}}, g^{3\frac{p(p-1)}{8}}, g^{5\frac{p(p-1)}{8}}, g^{7\frac{p(p-1)}{8}}.$ So if we find one such $n$ , then all $n$ are $n, n^3, n^5, n^7.$ Consider the $2$ from before. Note $17^2 \mid 2^{4 \cdot 17} + 1$ by LTE. Hence the possible $n$ are, $2^{17}, 2^{51}, 2^{85}, 2^{119}.$ Some modular arithmetic yields that $2^{51} \equiv \boxed{110}$ is the least value.

~Aaryabhatta1

Solution 4

These kinds of problems are, by nature, elementary. We get: $m^4 \equiv -1 \pmod{p^2},$ thus, $m$ is even, and $p^2 = 16k + 1$ or $p = 16k + 1,$ since $p$ is prime. Therefore, the smallest possible such $p$ is $17$ . Again, $m^4 \equiv 16 \pmod{17}.$ This is where it gets a bit tricky. $(m^2 - 4)(m^2 + 4) \equiv 0 \pmod{17}$ or $m^2 \equiv 13 \pmod{17}.$ This gives rise to: $m \equiv 8 \pmod{17}.$ Now, $m$ lies in the series $42, 76, 110, 144, \ldots$ . It is easy to see that the smallest value of $m$ is $110$ as neither $42$ nor $76$ satisfy all criteria.

~Grammaticus

Where is the justification for why $m$ is even? If there is none, then this is just a "lucky solve" ~inaccessibles

Video Solution

https://www.youtube.com/watch?v=_ambewDODiA

~MathProblemSolvingSkills.com

Video Solution 1 by OmegaLearn.org

https://youtu.be/UyoCHBeII6g

Video Solution 2

https://youtu.be/F3pezlR5WHc

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 6: / Line 6: @@
 If \(p=2\), then \(4\mid n^4+1\) for some integer \(n\). But \(\left(n^2\right)^2\equiv0\) or \(1\pmod4\), so it is impossible. Thus \(p\) is an odd prime.
-For integer \(n\) such that \(p^2\mid n^4+1\), we have \(p\mid n^4+1\), hence \(p\nmid n^4-1\), but \(p\mid n^8-1\). By Fermat's theorem, \(p\mid n^{p-1}-1\), so
+For integer \(n\) such that \(p^2\mid n^4+1\), we have \(p\mid n^4+1\), hence \(p\nmid n^4-1\), but \(p\mid n^8-1\). By [[Fermat's Little Theorem]], \(p\mid n^{p-1}-1\), so
 \begin{equation*}
 p\mid\gcd\left(n^{p-1}-1,n^8-1\right)=n^{\gcd(p-1,8)}-1.
@@ Line 15: / Line 15: @@
 \hline
 \vphantom{\tfrac11}x\bmod{17}&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\\hline
-\vphantom{\dfrac11}\left(x^4\right)^2+1\bmod{17}&2&0&14&2&14&5&5&0&0&5&5&14&2&14&0&2\\\hline
+\vphantom{\dfrac11}\left(x^4\right)+1\bmod{17}&2&0&14&2&14&5&5&0&0&5&5&14&2&14&0&2\\\hline
 \end{array}
 So \(m\equiv\pm2\), \(\pm8\pmod{17}\). If \(m\equiv2\pmod{17}\), let \(m=17k+2\), by the binomial theorem,
@@ Line 53: / Line 53: @@
 <cmath>\sqrt[4]{-1}=\pm\sqrt{\frac12}\pm\sqrt{-\frac12}.</cmath>
 Since \(-\frac12=144\), the expression becomes \(\pm12\pm12i\), and it is easily calculated via Hensel that \(i=38\), thus giving an answer of \(\boxed{110}\).
+==Solution 3 (Easy, given specialized knowledge)==
+Note that <math>n^4 + 1 \equiv 0 \pmod{p}</math> means <math>\text{ord}_{p}(n) = 8 \mid p-1.</math> The smallest prime that does this is <math>17</math> and <math>2^4 + 1 = 17</math> for example. Now let <math>g</math> be a primitive root of <math>17^2.</math> The satisfying <math>n</math> are of the form, <math>g^{\frac{p(p-1)}{8}}, g^{3\frac{p(p-1)}{8}}, g^{5\frac{p(p-1)}{8}}, g^{7\frac{p(p-1)}{8}}.</math> So if we find one such <math>n</math>, then all <math>n</math> are <math>n, n^3, n^5, n^7.</math> Consider the <math>2</math> from before. Note <math>17^2 \mid 2^{4 \cdot 17} + 1</math> by LTE. Hence the possible <math>n</math> are, <math>2^{17}, 2^{51}, 2^{85}, 2^{119}.</math> Some modular arithmetic yields that <math>2^{51} \equiv \boxed{110}</math> is the least value.
+~Aaryabhatta1
+==Solution 4==
+These kinds of problems are, by nature, elementary. We get: <cmath> m^4 \equiv -1 \pmod{p^2},</cmath> thus, <math>m</math> is even, and <cmath>p^2 = 16k + 1</cmath> or <cmath>p = 16k + 1,</cmath> since <math>p</math> is prime. Therefore, the smallest possible such <math>p</math> is <math>17</math>. Again,
+<cmath>m^4 \equiv 16 \pmod{17}.</cmath> This is where it gets a bit tricky.
+<cmath>(m^2 - 4)(m^2 + 4) \equiv 0 \pmod{17}</cmath> or <cmath>m^2 \equiv 13 \pmod{17}.</cmath> This gives rise to: <cmath>m \equiv 8 \pmod{17}.</cmath>
+Now, <math>m</math> lies in the series <math>42, 76, 110, 144, \ldots</math>. It is easy to see that the smallest value of <math>m</math> is <math>110</math> as neither <math>42</math> nor <math>76</math> satisfy all criteria.
+~Grammaticus
+Where is the justification for why <math>m</math> is even? If there is none, then this is just a "lucky solve"
+~inaccessibles
+==Video Solution==
+https://www.youtube.com/watch?v=_ambewDODiA
+~MathProblemSolvingSkills.com
 ==Video Solution 1 by OmegaLearn.org==

2024 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search