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

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 $p=13$.
