Clearly, $p \ge 3$. Suppose that there exist $a,b$ such that $a^2+2b^2=p^2 (1) \Rightarrow 2b^2=(p+a)(p-a) (2) $ Consider the $gcd(p+a,p-a)$ which divides $2p$ and must therefore be either $1,2,p,2p$ Case 1: $gcd(p+a,p-a)=p \Rightarrow p|a $which gives a contradiction as the LHS in $(1)$ will be larger than the right hand side. Case 2: $gcd(p+a,p-a)=2p \Rightarrow p|a$ which gives a contradiction as the LHS in $(1)$ will be larger than the right hand side. Case 3: $gcd(p+a,p-a)=1 \Rightarrow p+a, p-a$ are both odd, which means the RHS in $(2)$ is odd, while the LHS is even. Contradiction. Case 4: $gcd(p+a,p-a)=2 \Rightarrow b^2 = 2 \big( \frac{p+a}{2} \big) \big( \frac{p-a}{2} \big)$ so either $\frac{p+a}{2}=x^2, \frac{p-a}{2}=2y^2 \Rightarrow p=x^2+2y^2$ or $\frac{p+a}{2}=2x^2, \frac{p-a}{2}=y^2 \Rightarrow p=2x^2+y^2$ In either case, $p \in A$

Romania TST 1997.