Prove the modulo 12 pattern of $400*k^2 + 100k + 3$

Question

Consider the formula $$400*k^2 + 100k + 3$$ where k are whole positive integers.

The outcome of the modulo 12 values of x respectively seems to be: 11, 3, 3, 11, 3, 3 etc.

This question comes up while trying to prove some specific propery of odd triangular numbers on my own. I specifically need to prove that the outcome modulo 12 is never equal to 1.

Answer 1

Suppose that $400k^2+100k+3 \equiv 1 \ ({\sf mod} \ 12)$. Then $400k^2+100k+2 \equiv 0 \ ({\sf mod} \ 12)$, or $200k^2+50k+1 \equiv 0 \ ({\sf mod} \ 6)$, which means that the odd number $200k^2+50k+1$ is a multiple of $6$ which is impossible.

Answer 2

$\!\bmod 4\!:\ a_k = 400k^2\!+\!100k\!+\!3\equiv\color{#c00}{3},\,$ but $\bmod 12\!:\ a_k\equiv 1\,\Rightarrow \bmod 4\!:\ a_k\equiv\color{#c00}{ 1},\,$ contradiction.

Remark $\ $ We can also show $\bmod 12\!:\ a_k\equiv 3,11\,$ as you observed, namely

$\!\bmod 12\!:\,\ a_k \equiv 4k^2\!+\!4k\!+\!3\equiv 3,11\iff a_k+1 \equiv 4(k^2\!+\!k\!+\!1)\equiv 4,12$

$\begin{align}\bmod 3\!:\ k\equiv 0,2\,&\Rightarrow\, k^2\!+\!k\!+\!1\equiv 1\ \Rightarrow\,\bmod 12\!:\ a_k+1\equiv 4\,\Rightarrow\, a_k\equiv\ \ 3\\ \bmod 3\!:\ k\equiv 1\ \ \ \ \,& \Rightarrow\ k^2\!+\!k\!+\!1\equiv 0\,\Rightarrow\,\bmod 12\!:\ a_k+1\equiv 0\,\Rightarrow\,a_k\equiv -1\end{align}$