Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.
Impossible to be \(5q+2\) or \(5q+3\).
Step 1: Express an integer in terms of division by 5.
When any integer \(n\) is divided by 5, the remainder can only be 0, 1, 2, 3, or 4. So, every integer can be written in one of the following forms:
\(n = 5q\), or \(n = 5q + 1\), or \(n = 5q + 2\), or \(n = 5q + 3\), or \(n = 5q + 4\).
Step 2: Square each possibility.
If \(n = 5q\):
\(n^2 = (5q)^2 = 25q^2 = 5(5q^2)\).
This is clearly a multiple of 5, so the remainder is 0 when divided by 5.
If \(n = 5q + 1\):
\(n^2 = (5q + 1)^2 = 25q^2 + 10q + 1 = 5(5q^2 + 2q) + 1\).
So the remainder is 1 when divided by 5.
If \(n = 5q + 2\):
\(n^2 = (5q + 2)^2 = 25q^2 + 20q + 4 = 5(5q^2 + 4q) + 4\).
So the remainder is 4 when divided by 5.
If \(n = 5q + 3\):
\(n^2 = (5q + 3)^2 = 25q^2 + 30q + 9 = 25q^2 + 30q + 5 + 4\).
= \(5(5q^2 + 6q + 1) + 4\).
So the remainder is also 4 when divided by 5.
If \(n = 5q + 4\):
\(n^2 = (5q + 4)^2 = 25q^2 + 40q + 16 = 25q^2 + 40q + 15 + 1\).
= \(5(5q^2 + 8q + 3) + 1\).
So the remainder is 1 when divided by 5.
Step 3: Collect the results.
Therefore, depending on the form of \(n\), the possible remainders of \(n^2\) when divided by 5 are only:
0, 1, or 4.
Step 4: Compare with the given forms.
If \(n^2\) were of the form \(5q + 2\), its remainder upon division by 5 would be 2. Similarly, if \(n^2\) were of the form \(5q + 3\), its remainder would be 3.
But we have just shown that the remainder can only be 0, 1, or 4.
Conclusion. Hence, the square of any positive integer can never be of the form \(5q + 2\) or \(5q + 3\).