Show that the square of any positive integer is either of the form \(4q\) or \(4q + 1\) for some integer \(q\).
Squares are of the form \(4q\) or \(4q+1\).
Step 1: Consider a positive integer.
Let \(n\) be any positive integer. Every integer is either even or odd.
Step 2: Case when \(n\) is even.
If \(n\) is even, we can write \(n = 2k\) for some integer \(k\).
Then,
\(n^2 = (2k)^2\)
\(= 4k^2\)
This is a multiple of 4, so it is of the form \(4q\), where \(q = k^2\).
Step 3: Case when \(n\) is odd.
If \(n\) is odd, we can write \(n = 2k + 1\) for some integer \(k\).
Then,
\(n^2 = (2k + 1)^2\)
\(= 4k^2 + 4k + 1\)
\(= 4(k^2 + k) + 1\)
This is of the form \(4q + 1\), where \(q = k^2 + k\).
Step 4: Conclusion.
Therefore, the square of any positive integer is either of the form \(4q\) (if the number is even) or \(4q+1\) (if the number is odd).