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If x ≡ 0 (mod 4), then 3^x ≡ 3^0 ≡ 1 (mod 5)

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3^4 = 81 ≡ 1 (mod 5)

So 3^(4n) = (3^(4))^n ≡1^n (mod 5)≡ 1 (mod 5)

Simple as that. It's simply a true statement, nothing profound here.
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To add a little flavor to the other comment, write down the powers of 3, starting with 3^0 = 1:

1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, …

The units digit cycles through 1, 3, 9, 7 and then starts over again. So, you could say that when x=0 (4), 3^x = 1 (10) and when x=1 (4), 3^x = 3 (10), and so on for the 4 different values of x mod 4. But if 3^x is 1 mod 10, then 3^x is also 1 mod 5, and that’s what they are using here.
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> If x is 0 mod 4, then would it neccessarily be 0 mod 5?

You may want to research what mod means.   While 0 is 0 in any modulus, 4 is 0 mod 4 but not 0 mod 5.

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