# Difference between revisions of "2019 AIME I Problems/Problem 14"

## Problem 14

Find the least odd prime factor of $2019^8+1$.

## Solution

We know that $2019^8 \equiv -1 \pmod{p}$ for some prime $p$. We want to find the smallest odd possible value of $p$. By squaring both sides of the congruence, we find $2019^{16} \equiv 1 \pmod{p}$.

Since $2019^{16} \equiv 1 \pmod{p}$, the order of $2019$ modulo $p$ is a positive divisor of $16$.

However, if the order of $2019$ modulo $p$ is $1, 2, 4,$ or $8,$ then $2019^8$ will be equivalent to $1 \pmod{p},$ which contradicts the given requirement that $2019^8\equiv -1\pmod{p}$.

Therefore, the order of $2019$ modulo $p$ is $16$. Because all orders modulo $p$ divide $\phi(p)$, we see that $\phi(p)$ is a multiple of $16$. As $p$ is prime, $\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1$. Therefore, $p\equiv 1 \pmod{16}$. The two smallest primes equivalent to $1 \pmod{16}$ are $17$ and $97$. As $2019^8 \not\equiv -1 \pmod{17}$ and $2019^8 \equiv -1 \pmod{97}$, the smallest possible $p$ is thus $\boxed{097}$.

### Note to solution $\phi(k)$ is the Euler Totient Function of integer $k$. $\phi(k)$ is the number of positive integers less than $k$ relatively prime to $k$. Define the numbers $k_1,k_2,k_3,\cdots,k_n$ to be the prime factors of $k$. Then, we have $\phi(k)=k\cdot \prod^n_{i=1}\left(1-\dfrac{1}{k_i}\right).$ $A property of the Totient function is that, for any prime$p $,$\phi(p)=p-1$. [[Euler's Totient Theorem]] states that <cmath>a^{\phi(k)} \equiv 1\pmod k</cmath> if$ (Error compiling LaTeX. ! Missing $inserted.)\gcd(a,k)=1$.

Furthermore, the order$(Error compiling LaTeX. ! Missing$ inserted.)a $modulo$n $for an integer$a $relatively prime to$n $is defined as the smallest positive integer$d $such that$a^{d} \equiv 1\pmod n $. An important property of the order$d $is that$d|\phi(n)\$.

## Video Solution

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