Show that the square of any odd integer is of the form 4q + 1, for some integer q.
\(4q + 1\)
Step 1: Express an odd integer.
Any odd integer can be written as \(n = 2k + 1\), where \(k\) is some integer. This form ensures that \(n\) is always one more than an even number, hence odd.
Step 2: Square the expression.
Now square both sides:
\(n^2 = (2k + 1)^2\)
Step 3: Expand the square.
Using the formula \((a + b)^2 = a^2 + 2ab + b^2\), we get:
\(n^2 = (2k)^2 + 2 \cdot (2k) \cdot 1 + 1^2\)
\(n^2 = 4k^2 + 4k + 1\)
Step 4: Factorize.
Take 4 common from the first two terms:
\(n^2 = 4(k^2 + k) + 1\)
Step 5: Identify the form.
Let \(q = k^2 + k\), which is an integer (since \(k\) is an integer).
Thus, \(n^2 = 4q + 1\).
Conclusion. The square of any odd integer is always of the form \(4q + 1\) for some integer \(q\).