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

Latest revision as of 15:35, 27 September 2025

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.

There are two ways to do the first part:

First Method using FLT. 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 $p=17=2\times8+1$.

Second Method using orders. Notice that the order modulo $p^2$ of $n^4$ must be $8$ , so $8\mid\varphi(p^2)=p(p-1)$ . Hence $8\mid p-1$ , so $p\equiv 1\pmod{8}$ . The smallest such prime is $p=17$ . ~eevee9406

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 (Hensel's Lemma)

We know, from Solution 1, that by Fermat's Little Theorem, $p\equiv 1\pmod{8}$ , and that the smallest $p$ satisfying such is $17$ . Thus, we have $p^2=289$ and we are attempting to lift modulo $17$ to modulo $289$ .

Given in the problem, it is established that $p^2\mid m^4 +1$ holds true, so we can guarantee that $m^4\equiv 1\pmod{289}$ . Testing for roots $r_0$ modulo $17$ gives us the solutions $r_0= \pm 2$ , $\pm 8$ , which satisfy $r_0^4\equiv -1 \pmod{17}$ .

Note that we can now use Hensel's Lemma, by defining a function $f(x)=x^4+1$ , giving $f'(x)=4x^3$ . We know firstly that, though $f(r_0)$ for all roots $r_0$ we found is congruent to $0\pmod{17}$ , this does not hold true for $f'(r_0)$ , so we guarantee that there is one lift by Hensel's Lemma to $289$ that we can perform, defined for root $r_1$ that $r_1 = r_0-\frac{f(r_0)}{f'(r_0)}\pmod{p^k}$ . We now break our work into four cases:

$r_0 = 2$ . $f(2)=17$ and $f'(2)=32$ , so $r_1=2-\frac{17}{32}\pmod{289}$ , and we need to find the inverse of $32\equiv 15$ modulo $17$ . By the extended Euclidean algorithm, we find that $8\cdot 15\equiv 1 \pmod{17}$ , and thus $r_1=2-17\cdot 8\equiv 155 \pmod{289}$ .
$r_0 = -2$ , or equivalent, $15$ . $f(-2)=17$ and $f'(-2)=-32$ , so $r_1=-2+\frac{17}{32}\pmod{289}$ . We already found the inverse of $32$ modulo $17$ , so this case has $r_1=-1+17\cdot 8\equiv 134 \pmod{289}$ .
$r_0 = 8$ . $f(8)=4097$ and $f'(8)=2048$ , so $r_1=-2+\frac{4097}{2048}\pmod{289}$ , and we need to find the inverse of $32\equiv 8$ modulo $17$ . We already know the inverse of $8$ is $15$ , so we end up with $r_1=8-4097\cdot 15\equiv 110\pmod{289}$ .
$r_0 = -8$ , or equivalent, $9$ . $f(-8)=4097$ and $f'(-8)=-2048$ , so $r_1=-8+\frac{4097}{2048}\pmod{289}$ . We already found the inverse of $2048$ modulo $17$ , so this case has $r_1=-8+4097\cdot 15\equiv 179\pmod{289}$ .

Thus, out of our four values for $m$ , the smallest is $\boxed{110}$ . Solution by juwushu.

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 (Unfinished)

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 $\boxed{110}$.

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 2: / Line 2: @@
 Let <math>p</math> be the least prime number for which there exists a positive integer <math>n</math> such that <math>n^{4}+1</math> is divisible by <math>p^{2}</math>. Find the least positive integer <math>m</math> such that <math>m^{4}+1</math> is divisible by <math>p^{2}</math>.
-==Solution 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 theorem, \(p\mid n^{p-1}-1\), so
+There are two ways to do the first part:
+<b>First Method using FLT.</b> 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 \(p=17=2\times8+1\).
+<b>Second Method using orders.</b> Notice that the order modulo <math>p^2</math> of <math>n^4</math> must be <math>8</math>, so <math>8\mid\varphi(p^2)=p(p-1)</math>. Hence <math>8\mid p-1</math>, so <math>p\equiv 1\pmod{8}</math>. The smallest such prime is <math>p=17</math>. ~eevee9406
 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)^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 49: / Line 56: @@
 <font size=2>Solution by Quantum-Phantom</font>
-==Solution 3==
+==Solution 2 (Hensel's Lemma)==
+We know, from Solution 1, that by Fermat's Little Theorem, <math>p\equiv 1\pmod{8}</math>, and that the smallest <math>p</math> satisfying such is <math>17</math>. Thus, we have <math>p^2=289</math> and we are attempting to lift modulo <math>17</math> to modulo <math>289</math>.
+Given in the problem, it is established that <math>p^2\mid m^4 +1</math> holds true, so we can guarantee that <math>m^4\equiv 1\pmod{289}</math>. Testing for roots <math>r_0</math> modulo <math>17</math> gives us the solutions <math>r_0= \pm 2</math>, <math>\pm 8</math>, which satisfy <math>r_0^4\equiv -1 \pmod{17}</math>.
+Note that we can now use Hensel's Lemma, by defining a function <math>f(x)=x^4+1</math>, giving <math>f'(x)=4x^3</math>. We know firstly that, though <math>f(r_0)</math> for all roots <math>r_0</math> we found is congruent to <math>0\pmod{17}</math>, this does not hold true for <math>f'(r_0)</math>, so we guarantee that there is one lift by Hensel's Lemma to <math>289</math> that we can perform, defined for root <math>r_1</math> that <math>r_1 = r_0-\frac{f(r_0)}{f'(r_0)}\pmod{p^k}</math>. We now break our work into four cases:
+# <math>r_0 = 2</math>. <math>f(2)=17</math> and <math>f'(2)=32</math>, so <math>r_1=2-\frac{17}{32}\pmod{289}</math>, and we need to find the inverse of <math>32\equiv 15</math> modulo <math>17</math>. By the extended Euclidean algorithm, we find that <math>8\cdot 15\equiv 1 \pmod{17}</math>, and thus <math>r_1=2-17\cdot 8\equiv 155 \pmod{289}</math>.
+# <math>r_0 = -2</math>, or equivalent, <math>15</math>. <math>f(-2)=17</math> and <math>f'(-2)=-32</math>, so <math>r_1=-2+\frac{17}{32}\pmod{289}</math>. We already found the inverse of <math>32</math> modulo <math>17</math>, so this case has <math>r_1=-1+17\cdot 8\equiv 134 \pmod{289}</math>.
+# <math>r_0 = 8</math>. <math>f(8)=4097</math> and <math>f'(8)=2048</math>, so <math>r_1=-2+\frac{4097}{2048}\pmod{289}</math>, and we need to find the inverse of <math>32\equiv 8</math> modulo <math>17</math>. We already know the inverse of <math>8</math> is <math>15</math>, so we end up with <math>r_1=8-4097\cdot 15\equiv 110\pmod{289}</math>.
+# <math>r_0 = -8</math>, or equivalent, <math>9</math>. <math>f(-8)=4097</math> and <math>f'(-8)=-2048</math>, so <math>r_1=-8+\frac{4097}{2048}\pmod{289}</math>. We already found the inverse of <math>2048</math> modulo <math>17</math>, so this case has <math>r_1=-8+4097\cdot 15\equiv 179\pmod{289}</math>.
+Thus, out of our four values for <math>m</math>, the smallest is <math>\boxed{110}</math>. Solution by [[User:Juwushu|juwushu]].
+==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 (Unfinished)==
 We work in the ring \(\mathbb Z/289\mathbb Z\) and use the formula
 <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 \(\boxed{110.00}\).
+Since \(-\frac12=144\), the expression becomes \(\pm12\pm12i\), and it is easily calculated via Hensel that \(i=38\), thus giving an answer of \(\boxed{110}\).
+==Video Solution==
+https://www.youtube.com/watch?v=_ambewDODiA
+~MathProblemSolvingSkills.com
 ==Video Solution 1 by OmegaLearn.org==
-https://youtube/UyoCHBeII6g
+https://youtu.be/UyoCHBeII6g
 ==Video Solution 2==

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