Final answer:
The remainders for the given powers are 1 when divided by 2023 and -1 (or 2021) when divided by 2022. The square of any integer is congruent to 0, 1, or 4 mod 8. If 3 does not divide a, then 3 does not divide a², which helps prove √3 is irrational. The provided equivalences can be verified by modular arithmetic.
Step-by-step explanation:
The remainder when 2022²ˣ is divided by 2023 can be solved using the concept of modular arithmetic. Since 2022 is 1 less than 2023, we can express that 2022 ≈ -1 (mod 2023). Raising this to the power of 2022 gives us (-1)ˣ, which is 1 (mod 2023), as even powers of -1 are equal to 1. Thus, the remainder is 1.
Similarly, the remainder when 2021ˠˡ is divided by 2022 can be found using the same concept. Since 2021 is 1 less than 2022, we have 2021 ≈ -1 (mod 2022). Raising this to an odd power, 2021, results in -1, so the remainder is 2021.
For proving that for all integers n, n² ≃ 0mod8, 1mod8, or 4 mod8, we analyze the square of integers and their residues modulo 8. All integers can be expressed as 8k, 8k+1, 8k+2, ... 8k+7 for some integer k, and squaring each form can show that the square is congruent to either 0, 1, or 4 modulo 8.
To prove that if 3 does not divide a, then 3 does not divide a² can be shown by contrapositive: if 3 divides a², then it must divide a. This statement helps in demonstrating that √3 is an irrational number, as assuming √3 were rational and expressing it as a fraction of two integers where 3 does not divide the numerator leads to a contradiction, implying that √3 cannot be expressed as such a fraction.
The equivalences x²+x+3 ≃ 0mod5, (a-x²) ≃ x²moda, and x³+1 ≃ 0mod3 can be proved by evaluating each respective expression for different x values and checking their residues modulo the respective moduli.