Prove that if n is an odd positive integer, then 1 ≡ n2 (mod 4).
n = 2k+1
n^2 = 4k^2+4k+1 = 4(k^2+k) + 1
To prove that if n is an odd positive integer, then 1 ≡ n^2 (mod 4), we need to show that n^2 is congruent to 1 modulo 4.
Let's break down the process into smaller steps:
Step 1: Express n as an odd positive integer.
Since n is given to be an odd positive integer, we can write it as n = 2k + 1, where k is a non-negative integer.
Step 2: Substitute n from Step 1 into the equation n^2.
n^2 = (2k + 1)^2 = 4k^2 + 4k + 1
Step 3: Simplify the expression obtained in Step 2.
We can simplify the expression by factoring out 4 from the first two terms:
n^2 = 4(k^2 + k) + 1
Step 4: Observe the expression obtained in Step 3 and understand its properties.
Since k^2 + k is an integer, let's denote it as m, where m = k^2 + k.
So we have n^2 = 4m + 1.
Step 5: Determine the congruence of n^2 (mod 4).
To find the congruence of n^2 (mod 4), we need to divide n^2 by 4 and observe the remainder.
Divide n^2 by 4:
n^2 = 4m + 1 = 4m + (4 - 3) = 4m + 4 - 3 = 4(m + 1) - 3
The last expression, 4(m + 1) - 3, can be rewritten as:
4(m + 1) ≡ 3 (mod 4)
Step 6: Observe the final expression.
In the expression 4(m + 1) ≡ 3 (mod 4), we can see that the left side, 4(m + 1), is divisible by 4 without any remainders. Therefore, it is congruent to 0 mod 4.
So we have:
4(m + 1) ≡ 0 (mod 4)
Now, let's compare this to the right side: 3 (mod 4).
When we divide 3 by 4, we get a remainder of 3.
Therefore, we have proved that 4(m + 1) is congruent to 3 (mod 4):
4(m + 1) ≡ 3 (mod 4)
Step 7: Final conclusion.
Since 4(m + 1) ≡ 3 (mod 4), this implies that n^2 ≡ 1 (mod 4). Therefore, if n is an odd positive integer, then 1 ≡ n^2 (mod 4), as required.
Note: It's important to understand the properties of congruence and divisibility to follow this proof.