Perfect Square Patterns

A neat solution: All numbers can be written as 3k, 3k + 1, or 3k + 2, where k is a integer. Square each of these and find the terms not divisible by 3.

Case 1. 3k (3k)^2 = 9k^2 -> Remainder will be 0.

Case 2. 3k + 1 (3k+1)^2 = 9k^2 + 6k + 1 -> Remainder will be 1.

Case 3. 3k + 2 (3k+2)^2 = 9k^2 + 12k + 4 -> Remainder will be 4, or 1.

I just squared 3, 6, 9... et cetera, and saw they're all divisible by three with a remainder of zero.

Let $n$ be a perfect square. Assume $n\equiv 2\pmod{3}$ . We have:

$n\equiv 0^{2},1^{2},\textrm{or }2^{2} \pmod{3}$ (this holds true since $n$ is a perfect square)

$n\equiv 0,1,\textrm{or }4 \pmod{3}$

$0$ and $1$ check out. However,

$n\equiv 4 \pmod{3}$

$\implies n\equiv 1 \pmod{3}$

which contradicts the original assumption $n\equiv 2 \pmod{3}$ . Therefore, $n$ cannot be a perfect square with remainder 2. So, a perfect square may only have remainder $0$ or $1$ when divided by 3.