Power of primes We have proved that if $a > 3$, then $a$, $a+ 2$, and $a+ 4$ cannot be all primes in previous question. Can we say that 
they all be powers of primes?
 A: If you count first powers as powers of primes, you have $5,7,9$ and $7,9,11$ and $9,11,13$ and $23, 25, 27$ and $25,27,29$.
You can't have even numbers, and one will be a multiple of $3$, so will have to be a power of $3$, and this reduces the search space somewhat. $79, 81, 83$ is the next example.
I've no idea about the distribution of such things through.
A: I have requested Exponential diophantine equations by T.N. Shorey and R. Tijdeman from a nearby library as i am now curious. 
In case the requirement is for all exponents larger than one, we can probably find all solutions (likely none). The point is that one of the numbers is divisible by 3, therefore is $3^k.$ With unknown primes $p,q,r,s,$ exponents $t,u,v,w \geq 2,$ we want all solutions to
$$ 3^k + 2 = p^t,  $$ 
$$ 3^k + 4 = q^u,  $$
$$ 3^k - 2 = r^v,  $$
$$ 3^k - 4 = s^w.  $$
My guess is that, because of the fixed 3, these four can be solved individually.
Komputer Kalkulation: I just ran it up to $3^{20} \pm 2,4,$ all I found was $27 - 2 = 25.$ I suspect that is all...
Note that replacing the fixed $3$ by another unknown prime leads to open problems.
