Write whether the square of any positive integer can be of the form \(3m+2\), where \(m\) is a natural number. Justify your answer.
No.
Step 1: Expressing any integer in terms of 3.
Every integer \(n\) can be written in one of the following forms:
\(n = 3q\), or \(n = 3q + 1\), or \(n = 3q + 2\),
where \(q\) is some integer (quotient on division by 3).
Step 2: Squaring each case.
Case A: If \(n = 3q\)
Then \(n^2 = (3q)^2 = 9q^2\).
This can be written as \(n^2 = 3(3q^2)\), i.e. of the form \(3m\).
Case B: If \(n = 3q + 1\)
Then \(n^2 = (3q + 1)^2\).
Expanding: \(n^2 = 9q^2 + 6q + 1\).
This is \(n^2 = 3(3q^2 + 2q) + 1\).
So \(n^2\) is of the form \(3m + 1\).
Case C: If \(n = 3q + 2\)
Then \(n^2 = (3q + 2)^2\).
Expanding: \(n^2 = 9q^2 + 12q + 4\).
This can be written as \(n^2 = 3(3q^2 + 4q + 1) + 1\).
So \(n^2\) is again of the form \(3m + 1\).
Step 3: Combine results.
Thus, a square of any integer can only be of the form:
It can never be of the form \(3m + 2\).
Conclusion: The square of any positive integer cannot be of the form \(3m + 2\).