Does every odd prime determine a prime in an interval of length $\sqrt{p-2}$?

Let $p \geq 3$ be an odd prime. I would like to know whether the following conjecture is true.

Conjecture

For every odd prime $p \geq 3$, there exist integers $a$ and $b$ such that:

1. $a+b+3$ is prime;
2. $4a+2b+3=p$
3. $\gcd(a,b,3)=1$
4. $b^2\leq 4a$
5. $a\geq 1$.

Here, $\mathbb{P}$ denotes the set of prime numbers.

For example: all $$(a,b,p) \in \{(4,-4,3),(1,-1,5),(2,-2,7),(2,0,11)\}$$

work.

In each case, all five conditions are satisfied.

Equivalent formulation as a prime in a short interval

Set

$$ q:=a+b+3. $$

Then $q$ is prime. Solving the system

$$ q=a+b+3 $$

and

$$ p=4a+2b+3 $$

for $a$ and $b$, we obtain

$$ a=\frac{p-2q+3}{2} $$

and

$$ b=\frac{4q-p-9}{2}. $$

Since $p$ and $q$ are odd primes, both $a$ and $b$ are integers.

The inequality

$$ b^2\leq 4a $$

is equivalent to

$$ \left(\frac{4q-p-9}{2}\right)^2 \leq 4\left(\frac{p-2q+3}{2}\right). $$

Multiplying by $4$ gives

$$ \left(4q-p-9\right)^2 \leq 8\left(p-2q+3\right). $$

After simplifying, this becomes

$$ \left(4q-p-7\right)^2 \leq 4\left(p-2\right). $$

Therefore,

$$ -2\sqrt{p-2} \leq 4q-p-7 \leq 2\sqrt{p-2}. $$

Equivalently,

$$ \frac{p+7-2\sqrt{p-2}}{4} \leq q \leq \frac{p+7+2\sqrt{p-2}}{4}. $$

Define

$$ L(p):= \frac{p+7-2\sqrt{p-2}}{4} $$

and

$$ U(p):= \frac{p+7+2\sqrt{p-2}}{4}. $$

The length of this interval is

$$ U(p)-L(p)=\sqrt{p-2}, $$

and its midpoint is

$$ \frac{p+7}{4}. $$

Thus, the conjecture is essentially equivalent to the following statement:

For every odd prime $p \geq 3$, the interval

$$ \left[ \frac{p+7-2\sqrt{p-2}}{4}, \frac{p+7+2\sqrt{p-2}}{4} \right] $$ contains a prime.

For sufficiently large $p$, the condition $a\geq 1$ is automatic. Indeed,

$$ a=\frac{p-2q+3}{2}\geq 1 $$

is equivalent to

$$ q\leq \frac{p+1}{2}. $$

For $p\geq 11$, one has

$$ U(p)\leq \frac{p+1}{2}, $$

so every prime $q\in[L(p),U(p)]$ automatically gives $a\geq 1$. The cases $p=3,5,7$ can be checked directly.

The condition

$$ \gcd(a,b,3)=1 $$

is also automatic. If $3\mid a$ and $3\mid b$, then

$$ p=4a+2b+3 $$

and

$$ q=a+b+3 $$

would both be divisible by $3$. Since both are prime, this would force

$$ p=q=3. $$

But then

$$ a+b=0 $$

and

$$ 4a+2b=0. $$

Subtracting twice the first equation from the second gives

$$ 2a=0, $$

so $a=0$, contradicting $a\geq 1$.

Heuristic

The interval is centered near $p/4$ and has length approximately $\sqrt{p}$. The prime number theorem therefore suggests that the expected number of primes in the interval is approximately

$$ \frac{\sqrt{p}}{\log p}, $$

which tends to infinity as $p\to\infty$.

Cramér's conjecture would imply that every sufficiently large interval of this form contains a prime because

$$ \left(\log x\right)^2=o\left(\sqrt{x}\right). $$

However, known unconditional bounds for primes in short intervals do not appear to guarantee a prime in every interval of length comparable to $\sqrt{x}$.

Questions

1. Is this conjecture already known?

2. Can one prove unconditionally that the interval

   $$ [L(p),U(p)] $$

   contains a prime for every odd prime $p$?

3. Does the fact that $p$ itself is prime make the problem easier than the general problem of finding primes in every interval of length comparable to $\sqrt{x}$?

4. Does this conjecture follow from a standard conjecture weaker than Cramér's conjecture?

Important: What it needs to be proven is that for all primes $p \geq 3$ the interval $[L(p), U(p)]$ contains at least one prime. However, I proved that there are infinitely many primes $p \geq 3$ such that the interval contains at least one prime in $[L(p), U(p)]$.