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

Latest revision as of 17:43, 2 March 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)^2+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

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)

2024 AIME I (Problems • Answer Key • Resources)
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