2019 AIME I Problems/Problem 14

The 2019 AIME I takes place on March 13, 2019.

Problem 14

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

Solution 2 the solution in the chat is better and should be sol 1

Essentially, we are trying to find the smallest prime p such that $2019^8 \equiv -1 (\mod p)$ . This congruence tells us that $2019^{16} \equiv 1 (\mod p)$ . Therefore, the order of 2019 modulo p is a divisor of 16, but not a divisor of 8. This tells us that the order of 2019 modulo p is exactly 16 since it is the only possibility. We know that the order of 2019 modulo p is a divisor of $\phi(p)$ , which is just p-1 because p is prime.Thus, we have $16|p-1$ . Now, we just test up. 17 does not work, because $2019^8 +1$ reduces to 2 modulo 17. The reason this does not work is because $2019^8$ reduces to 1, not -1 modulo 17. Since, 33, 49, 65, and 81 are all composite, we are able to skip those cases. This brings us to 97, which gives $2019^8 \equiv (-18)^8 \equiv 324^4 \equiv 33^4 \equiv 1089^2 \equiv 22^2 \equiv 96 (\mod 97)$ . Thus $2019^8 +1 \equiv 0 (\mod\boxed{97})$ .

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2019 AIME I Problems/Problem 14

Problem 14

Solution 2 the solution in the chat is better and should be sol 1

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