Proof that number ending in $66$ or $06$ is congruent to $2 \text{ mod } 4$

Question

Let $N$ be an integer such that we can represent $N$ by its digits as $N=a_na_{n-1}\cdots a_1a_0$. We want to prove that if $a_0,a_1=6$ or if $a_1=0$ and $a_0=6$ then $N\equiv 2 \text{ mod } 4$.

I'm unsure how to proceed here.

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I also want to prove that $N$ is a perfect square only if $N\equiv 0 \text{ mod} 4$ or $N\equiv 1 \text{ mod} 4$.

My attempt Through trial with squares $1,2,3$ I notice a cyclic pattern $1,0,1,0...$ as the remainder. I proceeded by induction assuming that $k^2 \equiv 0 \text{ mod} 4$. thus $$k^2+1 \equiv 1 \text{ mod} 4$$ But I'm unsure how to get that $2k\equiv 0 \text{ mod } 4$ to complete this. Is the exponentiation identity $a\equiv b \text{ mod } c$ implies that $a^2\equiv b^2 \text{ mod } c$ an iff statement? If so my induction hypothesis gives that $k \equiv 0 \text{ mod } 4$ and the mulitplication rule gives $2k$ congruent to $0$, where I can then apply the addition rule.

Answer 1

These numbers are $100n+66$ or $100n+6$ since $100=0$ mod $4$ and $66=2$ mod $4$ ($6=2$ mod $4$) the result follows.

Answer 2

part 1:
Any $N$ ending in the digits $66$ or $06$ will be of the form $N=20k+6$.

Then since $20k\equiv 0 \bmod 4$ and $6\equiv 2\bmod 4$ we will have $N\equiv 2\bmod 4$.

part 2:
Suppose $N$ is a perfect square. Either $N$ is an odd square or an even square.

Even: $N=(2k)^2=4k^2\equiv 0 \bmod 4$

Odd: $N=(2k+1)^2=4k^2+4k+1\equiv 1 \bmod 4$

Thus $N$ can only be a perfect square if $N \in \{0,1\} \bmod 4$