The New Year Coincidence?

The following are valid:

201 9 2020 m o d 31 202 0 2019 m o d 31 1 2019^{2020} \bmod 31 \equiv 2020^{2019} \bmod 31 \equiv 1

202 0 2021 m o d 31 202 1 2020 m o d 31 25 2020^{2021} \bmod 31 \equiv 2021^{2020} \bmod 31 \equiv 25

Must 202 1 2022 m o d 31 202 2 2021 m o d 31 2021^{2022} \bmod 31 \equiv 2022^{2021} \bmod 31 ?

Yes. No.

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1 solution

Sathvik Acharya
Dec 31, 2020

Since we are given that 202 1 2020 25 ( m o d 31 ) 2021^{2020}\equiv 25 \pmod{31} and 2021 6 ( m o d 31 ) 2021\equiv 6 \pmod{31} , 202 1 2022 202 1 2020 202 1 2 25 6 2 900 1 ( m o d 31 ) 2021^{2022}\equiv 2021^{2020}\cdot 2021^{2}\equiv 25\cdot 6^2\equiv 900\equiv 1 \pmod{31} Using the fact that 2022 7 ( m o d 31 ) 2022\equiv 7\pmod{31} , 202 2 2021 7 2021 ( m o d 31 ) 2022^{2021}\equiv 7^{2021}\pmod{31} Note that 31 31 is prime. By Fermat's Little Theorem , we have, 7 30 1 ( m o d 31 ) 7^{30}\equiv 1\pmod{31} , 7 2021 ( 7 30 ) 67 7 11 7 11 ( 7 3 ) 3 7 2 2 3 7 2 20 ( m o d 31 ) 7^{2021}\equiv(7^{30})^{67}\cdot 7^{11}\equiv 7^{11}\equiv (7^3)^3\cdot 7^2\equiv 2^3\cdot 7^2\equiv 20 \pmod{31}

Therefore, 202 1 2022 1 ( m o d 31 ) 2021^{2022}\equiv 1\pmod{31} and 202 2 2021 20 ( m o d 31 ) 2022^{2021}\equiv 20\pmod{31} which implies that they are not equivalent.

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