Let be $p^2=a^2+2b^2$ with $b\neq 0 \implies p \nmid b$. Then looking $\mod p$ gives that $z^2 = -2 \mod p$ has a solution, namely $z=ab^{-1}$. Now take integers $x,y$ with $x \equiv zy \mod p$ and $|x|,|y| < \sqrt p, y \neq 0$ by Thue's lemma. But then $x^2+2y^2 \equiv 0 \mod p$ and $0< x^2+2y^2<3p$. If $x^2+2y^2=p$ we are done. In the other case $x^2+2y^2=2p$, $x$ is even, so $x=2x'$ and we get $y^2+2x'^2=p$.

Well in fact the more general requirment is true: Let $p$ is a prime. If $p$ is an divisor of an element in form $x^2+2y^2$ and $(p,y)=1$ then $p$ is in $A$. Here's in fact the same with ZetaX solution : assume that $a,b\in\mathbb Z$ which $b\neq0$ and $p\mid a^2+2b^2$ and $(p,b)=1$ we will find $c,d$ so that $p=c^2+2d^2$. If $p=2$ then chose $c=0$ and $d=1$. Assume that $p\geq3$. So we can find $x>0$ so that $x^2+2\equiv 0\pmod{p}$ (by mutiple the inverse of $d$ in modulo $p$.) Now consider all pair $(u,v)$ which $0\leq u,v\leq [\sqrt{p}]=k$ we have $(k+1)^2>p$ pair so have more than $p$ number in form $u+xv$ so we can find two distince pair $(u_i,v_i)$ which $i=1,2$ so that congurent to each other modulo $p$. Which gives us \[ u_1-u_2\equiv x(v_2-x_1)\pmod{p} \] Chose $c=|u_1-u_2|$ and $d=|v_1-v_2$ then $c,d\neq0$ and \[ c^2\equiv x^2d^2\equiv -2d^2\pmod{p} \] So $p\mid c^2+2d^2$. Now note that $0<2+2d^2<p+2p=3p$ so $c^2+2d^2=p$ or $2p$. If $c^2+2d^2=2p$ then $c$ is even and we chose $c_1=c/2, d_1=d/2$ we have $2c_1^2+d_1^2=p$. Of cause $c,c_1,d,d_1\neq0$ so we have $p$ must be longs to $A$.

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=470&t=150923 and http://www.artofproblemsolving.com/Forum/viewtopic.php?f=59&t=327

here there is realy an easier solution i think... let p^2=a^2+2b^2 => p^2-a^2=2b^2 =>(p-a)(p+a)=2b^2 and we know that p^2=1(mod 4) => b^2=0(mod 4) =>b=0(mod 2) and (p-a, p+a)=2 => p-a=2(x^2) and p+a=y^2 where x=1(mod 2) and y=0(mod 2) or p-a=x^2 and p+a=2(y^2) where x=0(mod 2) and y=1(mod2) in each of these two conditions we have: 2p=2w^2+z^2 which w=1(mod 2)and z=0(mod 2)(=>z^2= (mod 4)) =>p=w^2+(z^2)/2 and we had z^2=0(mod 4) so we could show that p is a member of A...