5= (1+4)

(1+4)^n= 1^n + 2*1*4 + ...+4^n

which is 1 mod 4

Prove that 5^n = 1 (mod 4) for any n >or= 1.

To prove that 5^n ≡ 1 (mod 4) for any n ≥ 1, we need to show that the remainder when dividing 5^n by 4 is always 1.

First, let's observe the pattern when we calculate the powers of 5 modulo 4:

5^1 ≡ 1 (mod 4)
5^2 ≡ 1 (mod 4)
5^3 ≡ 1 (mod 4)
...

Based on this pattern, we can hypothesize that 5^n ≡ 1 (mod 4) for any n ≥ 1. To prove this, we can use mathematical induction.

Step 1: Base case
Let's prove the statement for n = 1:
5^1 = 5, and the remainder when dividing 5 by 4 is 1. Therefore, the base case is true.

Step 2: Inductive step
Assume the statement is true for some positive integer k. That is, assume 5^k ≡ 1 (mod 4).

Now let's prove the statement for k + 1:
5^(k+1) = 5^k * 5
By the inductive hypothesis, we know that 5^k ≡ 1 (mod 4).
So, 5^(k+1) ≡ 1 * 5 ≡ 5 (mod 4).

We can rewrite 5 ≡ 1 (mod 4) as 1 ≡ 4 - 3 (mod 4).
Therefore, 5^(k+1) ≡ (4 - 3) ≡ 1 (mod 4).
In other words, the remainder when dividing 5^(k+1) by 4 is 1.

Since we have proven the base case (n = 1) and shown that if the statement is true for k, it is also true for k + 1, we can conclude that for any n ≥ 1, 5^n ≡ 1 (mod 4).

Hence, 5^n = 1 (mod 4) for any n ≥ 1 is proven.