Show that the square of any odd integer is of the form 4q+1, for some integer q.

To prove that the square of any odd integer is of the form 4q + 1, where q is some integer, let's break the problem into steps.

Step 1: Represent an odd integer
Any odd integer can be written in the form of 2n + 1, where n is some integer. This is because odd integers are those numbers that are not divisible by 2 and have a remainder of 1 when divided by 2.

Step 2: Square the odd integer
Now, square the odd integer 2n + 1.

(2n + 1)² = (2n)² + 2 * (2n) * 1 + 1²
= 4n² + 4n + 1

Step 3: Factor out 4
We can now factor out 4 from the first two terms:

4n² + 4n + 1 = 4(n² + n) + 1

Step 4: Identify the form of the result
Observe that the expression 4(n² + n) + 1 is of the form 4q + 1, where q = n² + n (which is an integer because n is an integer).

Conclusion:
Thus, the square of any odd integer is of the form 4q + 1, where q is an integer (specifically q = n² + n). This proves the statement.