# 2019 AIME I Problems/Problem 14

## Problem 14

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

## Solution

The problem tells us 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 get $2019^{16} \equiv 1 \pmod{p}$.

Since $2019^{16} \equiv 1 \pmod{p}$, $ord_p(2019)$ = $1, 2, 4, 8,$ or $16$

However, if $ord_p(2019)$ = $1, 2, 4,$ or $8,$ then $2019^8$ clearly will be $1 \pmod{p}$ instead of $-1 \pmod{p}$, causing a contradiction.

Therefore, $ord_p(2019) = 16$. Because $ord_p(2019) \vert \phi(p)$, $\phi(p)$ is a multiple of 16. Since we know $p$ is prime, $\phi(p) = p(1 - \frac{1}{p})$ or $p - 1$. Therefore, $p$ must be $1 \pmod{16}$. The two smallest primes that are $1 \pmod{16}$ are $17$ and $97$. $2019^8 \not\equiv -1 \pmod{17}$, but $2019^8 \equiv -1 \pmod{97}$, so our answer is $\boxed{97}$.

### Note to solution $\phi(p)$ is called the Euler Totient Function of integer $p$. Euler's Totient Theorem: define $\phi(p)$ as the number of positive integers less than $p$ but relatively prime to $p$, then we have $$\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})$$ where $p_1,p_2,...,p_n$ are the prime factors of $p$. Then, we have $$a^{\phi(p)} \equiv 1\ (\mathrm{mod}\ p)$$ if $\gcd(a,p)=1$.

Furthermore, $ord_n(a)$ for an integer $a$ relatively prime to $n$ is defined as the smallest positive integer $d$ such that $a^{d} \equiv 1\ (\mathrm{mod}\ n)$. An important property of the order is that $ord_n(a)|\phi(n)$.

## Video Solution

On The Spot STEM:

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