Question

Show that the square of an odd integer is of the form \( 8k + 1 \).

Hint:

To prove statements about odd integers, first write the odd number in the form \( 2n+1 \), then simplify step by step.

Answer 1

Step 1: Represent an odd integer in general form.}
Any odd integer can be written as \[ 2n + 1 \] where \( n \) is an integer.
Step 2: Square the odd integer.}
Now, squaring \( 2n + 1 \), we get \[ (2n + 1)^2 = 4n^2 + 4n + 1 \]
Step 3: Factor the expression.}
Taking \( 4n \) common from the first two terms, \[ 4n^2 + 4n + 1 = 4n(n+1) + 1 \]
Step 4: Use the property of consecutive integers.}
Here, \( n \) and \( n+1 \) are two consecutive integers. Out of two consecutive integers, one must be even. Therefore, the product \( n(n+1) \) is always even. So, we can write \[ n(n+1) = 2k \] for some integer \( k \).
Step 5: Substitute and obtain the required form.}
Substituting \( n(n+1) = 2k \) into the expression, we get \[ (2n+1)^2 = 4(2k) + 1 = 8k + 1 \] Hence, the square of an odd integer is always of the form \[ \boxed{8k + 1} \]