Finding primes $p$ such that $ \dfrac {p+1}2$ and $\dfrac{p-1}4$ are primes

Question

How many odd primes $p$ are there such that both $ \dfrac {p+1}2$ and $\dfrac{p-1}4$ are primes ?

Answer 1

One of $p,p+1,p+2$ is divisible by $3$, dividing by 2 or 4 doesn't change this so for all of these numbers to be prime we must have either $p=3$, $\frac{p+1}{2}=3$ or $\frac{p-1}{4}=3$, solve for $p$ and you'll find that $p=13$ is the only prime with the property you wanted

Answer 2

Think modulo 12. Either $p = 3$ or $p \cong \{1,5,7,11\} \mod 12$. $3$ doesn't work because $(3-1)/4$ isn't an integer. $\frac{\{1,5,7,11\}+1}{2} \cong \{1,3,4,6\} \mod 12$ so your $\frac{p+1}{2}$ prime requirement eliminates any $p \not \cong 1 \mod 12$. If $p \cong 1 \mod 12$ and $p>1$, then $12$ divides $p-1$, so $3$ divides $\frac{p-1}{4}$. If $\frac{p-1}{3\cdot4} = 1$ then we get the solution $p=13$. Otherwise some multiple of $3$ (other than just $3$) divides $\frac{p-1}{4}$, so it is not prime.

Therefore the only solution is $p=13$.