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 $a^{\phi(k)} \equiv 1\pmod k$ if $\gcd(a,k)=1$ .

Furthermore, the order $a$ 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

On The Spot STEM:

https://youtu.be/_vHq5_5qCd8

https://youtu.be/IF88iO5keFo

2019 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2019 AIME I Problems/Problem 14

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Problem 14

Solution

Note to solution

Video Solution

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