AIME 2024 I · 第 13 题

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:

1. $r_0 = 2$. $f(2)=17$ 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 $r_1=2-17\cdot 8\equiv 155 \pmod{289}$.
2. $r_0 = -2$, 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 $r_1=-1+17\cdot 8\equiv 134 \pmod{289}$.
3. $r_0 = 8$. $f(8)=4097$ 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 $r_1=8-4097\cdot 15\equiv 110\pmod{289}$.
4. $r_0 = -8$, 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)