How do I determine whether $18 \notin A$ with these premises? With $A \subseteq \mathbb{N}$, given that

A1. $(x \in A \land y \in A) \implies x^2 + y ^2 \in A$
A2. $1 \in A$
A3. $3 \notin A$

Determine whether:

$18 \notin A$

I've been unable to prove this. First, I tried to demonstrate $18 \in A$, but no matter what I try, I can't land in $18$, so I guess that $18 \notin A$ should be true - but how can I prove it? I have never used $3 \notin A$, which is probably the key, but I'm not sure how.
 A: Hint: (1) Give an example of a set $A$ that satisfies the conditions and such that $18\in A$.
(2) Give an example of a set $A$ that satisfies the conditions but such that $18\not\in A$. 
These two examples will show that (1) from the assumptions, we cannot prove that $18\not\in A$ and (2) from the assumptions we cannot prove that $18\in A$.
A: \begin{align}
\phantom{\left(1 \in A\quad \wedge\quad 1 \in A\right)} 
&\phantom{\quad\Longrightarrow\quad
1^{2} + 1^{2} =} 1 \in A 
\\
\left(1 \in A\quad \wedge\quad 1 \in A\right) 
&\quad\Longrightarrow\quad
1^{2} + 1^{2} = 2 \in A 
\\
\phantom{\left(1 \in A\quad \wedge\quad 1 \in A\right)} 
&\phantom{\quad\Longrightarrow\quad
1^{2} + 1^{2} =} \color{#ff0000}{\,3 \not\in A} 
\\
\left(1 \in A\quad \wedge\quad 2 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 2^{2} = 5 \in A 
\\
\left(2 \in A\quad \wedge\quad 2 \in A\right)
&\quad\Longrightarrow\quad
2^{2} + 2^{2} = 8 \in A 
\\
\left(1 \in A\quad \wedge\quad 3 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 3^{2} = 10 \in A 
\\
\left(2 \in A\quad \wedge\quad 3 \in A\right)
&\quad\Longrightarrow\quad
2^{2} + 3^{2} = 13 \in A 
\\
\left(1 \in A\quad \wedge\quad 4 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 4^{2} = 17 \in A 
\\
\left(4 \in A\quad \wedge\quad 2 \in A\right)
&\quad\Longrightarrow\quad
4^{2} + 2^{2} = 20 \in A 
\\
\left(1 \in A\quad \wedge\quad 5 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 5^{2} = 26 \in A 
\\
\left(2 \in A\quad \wedge\quad 5 \in A\right)
&\quad\Longrightarrow\quad
2^{2} + 5^{2} = 29 \in A 
\\
\left(1 \in A\quad \wedge\quad 6 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 6^{2} = 37 \in A 
\\
\left(2 \in A\quad \wedge\quad 6 \in A\right)
&\quad\Longrightarrow\quad
2^{2} + 6^{2} = 40 \in A 
\\
\left(4 \in A\quad \wedge\quad 5 \in A\right)
&\quad\Longrightarrow\quad
4^{2} + 5^{2} = 41 \in A 
\\&\vdots
\\[5mm]&
\end{align}
$$
\left\{%
1,2,5,8,10,13,17,20,26,29,37,40,41...
\right\}
$$
