Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.
Impossible to be \(6m + 2\) or \(6m + 5\).
Idea. Every integer, when divided by 6, leaves a remainder from \(0\) to \(5\). We will check the square in each case and look at the remainder (mod 6).
Step 1: Write the number by division algorithm.
Let \(n\) be any positive integer. Then \(n = 6q + r\).
Here \(q\) is an integer and \(r\in\{0,1,2,3,4,5\}\).
Step 2: Square in each remainder case and reduce modulo 6.
Case \(r=0\): \(n = 6q\), so \(n^2 = 36q^2\).
Thus \(n^2\equiv 0\pmod{6}\).
Case \(r=1\): \(n = 6q + 1\), so \(n^2 = 36q^2 + 12q + 1\).
Thus \(n^2\equiv 1\pmod{6}\).
Case \(r=2\): \(n = 6q + 2\), so \(n^2 = 36q^2 + 24q + 4\).
Thus \(n^2\equiv 4\pmod{6}\).
Case \(r=3\): \(n = 6q + 3\), so \(n^2 = 36q^2 + 36q + 9\).
Since \(9\equiv 3\pmod{6}\), we get \(n^2\equiv 3\pmod{6}\).
Case \(r=4\): \(n = 6q + 4\), so \(n^2 = 36q^2 + 48q + 16\).
Since \(16\equiv 4\pmod{6}\), we get \(n^2\equiv 4\pmod{6}\).
Case \(r=5\): \(n = 6q + 5\), so \(n^2 = 36q^2 + 60q + 25\).
Since \(25\equiv 1\pmod{6}\), we get \(n^2\equiv 1\pmod{6}\).
Step 3: Collect possible square remainders.
From the cases we have \(n^2\equiv 0, 1, 3, 4\pmod{6}\).
Therefore \(n^2\) can never have remainder \(2\) or \(5\) upon division by 6.
Conclusion. A perfect square cannot be of the form \(6m + 2\) or \(6m + 5\) for any integer \(m\).