Primes having 2 as a primitive root

Question

Let $p$ a prime and $a$ a quadratic residue of $p$. If there exists some positive integer $y$ satisfying $a\equiv2^y \pmod{p}$, prove or disprove that $2$ is a primitive root of $p$. My progress is that if 2 is not a primitive root then the set $\{2^0, 2^1, ..., 2^\frac{p-1}{2}\}$ is exactly the set of quadratic residues. Also $ord_p(2)=\frac{p-1}{2}$ or $p-1$.

Answer 1

As the comments point out whether $2$ is a primitive root of $p$ depends on $p$, $a$ and $y$, this answer is trying to figure out WHEN $2$ is a primitive root of $p$.

As $$a\equiv 2^y \pmod{p},$$ where $a$ is a quadratic residue of $p$ and $p$ is a prime number, it is clear that

Proposition 1: $2$ is NOT a primitive root of $p$ if $p=2$ or $y=1$.

In the following, assume that $p>2$ is an odd prime number and $2\le y \le p-2$. We assert that

Proposition: $2$ is NOT a primitive root of $p>2$ if $y$ is odd.

Proof: Let $g$ be a primitive root of $p$. Then there exist integers $1\le k \le p-1,0\le s\le \frac{p-1}{2}$ such that $$2=g^k,a=g^{2s}.$$ Then it follows from $a=2^y$ that $g^{2s}=g^{ky}$. Note that the order of $g$ is $p-1$. So we have $$ky\equiv 2s \pmod{p-1}.$$ Since $2s$ and $p-1$ are even, $ky$ must be even too. If $y$ is odd, then $k$ must be even, i.e., $2\mid k$. Then $$\mathrm{Ord}(2)=\frac{p-1}{\gcd(p-1,k)}=\frac{p-1}{2}\cdot \frac{1}{\gcd(\frac{p-1}{2},\frac{k}{2})}\le \frac{p-1}{2}.$$ Therefore, $2$ can be a primitive root of $p$. QED.

However, it is not a sufficient condition. After observing the result of the magma script below, a conjecture is obtained.

Conjecture 1: $2$ is a primitive root of $p$ if $y$ is even and $P(p)$, where $P(p)$ is a proposition of $p$ only.

We know whether $2$ is a primitive root of $p$ is completely determinated by $p$. But so far as we know, it is hard to give a YES or NO when only given $p$. So The proposition $P$ in Conjecture 1 is expected to be much more simple.

gp:={}; bp:={};
for p in [i: i in [3..200] | IsPrime(i)] do
base:=ResidueClassRing(p);
g:=PrimitiveRoot(p);
for y in [1..p-2] do
for s in [0..((p-1) div 2)] do
a:=g^(2*s);
if a eq (base!2)^y then
    flag:=IsPrimitive(base!2);
    if flag and not (y mod 2 eq 0) then
        print "bad1:", p,y,s,Order(base!2);
    end if;
    if flag and (y mod 2 eq 0) then
      gp join:= {p};
    end if;
    if not flag and (y mod 2 eq 0) then
      bp join:= {p};
    end if;
end if;

end for;
end for;
end for;

print bp meet gp;