Let $q = \frac{p-1}{2}$ and $r = \frac{p+1}{4}$; then $p = 4r-1$ and $q = \frac{4r-2}{2} = 2r-1$. Consider all remainders that can be left when $r$ is divided by $3$:
* If $r \equiv 1 \pmod{3}$ then $4r-1 \equiv 0 \pmod{3}$, i.e., $4r-1$ is divisible by $3$. Thus $p=3$. But then $r=1$ which is not a prime.
* If $r \equiv 2 \pmod{3}$ then $2r-1 \equiv 0 \pmod{3}$, i.e., $2r-1$ is divisible by $3$. Thus $q=3$, whence $r=2$ and $p=7$. All three are primes indeed.
* If $r \equiv 0 \pmod{3}$, i.e., $r$ is divisible by $3$, then $r=3$. Thus $q=5$ and $p=11$ which are primes, too.
Consequently, $p$ can be either $7$ or $11$.

Out of three consecutive integers $p-1$, $p$, $p+1$ one is divisible by $3$. Division by $2$ or $4$ does not change divisibility by $3$ since $2$ and $4$ are coprime with $3$. Thus also out of integers $p$, $\frac{p-1}{2}$, $\frac{p+1}{4}$ one is divisible by $3$. If these integers are prime, the one divisible by $3$ must be $3$. Consider all possible cases:
* If $p = 3$ then $\frac{p-1}{2} = 1$ but $1$ is not prime.
* If $\frac{p-1}{2} = 3$ then $p = 7$ and $\frac{p+1}{4} = 2$. All three are primes indeed.
* If $\frac{p+1}{4} = 3$ then $p = 11$ and $\frac{p-1}{2} = 5$. All three are primes indeed.
Consequently, $p$ can be either $7$ or $11$.