I found this elementary number theory problem in the “Problem Drive” section of Invariant Magazine (Issue 16, 2005), published by the Student Mathematical Society of the University of Oxford. Below, I have included the solution, which is very elementary.
Problem: Find all ordered pairs of prime numbers 
such that 
is also a prime.
Solution: Let 
. First, note that if is a solution, then so is 
. Now, 
and 
can’t be both even or both odd, else 
will be even. Without loss of generality, assume 
and some odd prime. So, 
. There are two cases to consider.
Case 1: 
.
This yields 
, which is prime. So, 
and, hence 
are solutions.
Case 2: 
.
There are two sub-cases to consider.
, where 
is some even integer. Then, we have 
. Hence, 
; so, can’t be prime.
, where is some odd integer. Then we have 
. Hence, ; so, again, can’t be prime.
As we have exhausted all possible cases, we conclude and are the only possible solutions.
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January 8, 2008 at 8:06 pm
John smith
but p,q are prime, so are odd by definition.
Can’t we just try (2,3) which is reasonable then consider all odd numbers p and q which contradicts the original assertion that p,q are prime since as you say p,q both odd means E is even.
So why do we need the rest of the working out you’ve done?
January 9, 2008 at 5:17 pm
Anonymous
Note that 2 is a prime.
And, there are four cases to consider for (p,q): (odd, odd), (odd, even), (even, even) and (even, odd). We have to discard the first and the third cases, since this makes E even. So, we are left with second and fourth cases. Since, E is symmetric with respect to p and q, we only need consider the fourth case, i.e. (p,q) is (even, odd). This forces p=2, and then we consider a couple of subcases to show that even this case fails.
January 9, 2008 at 5:19 pm
Vishal
The above was my comment actually.
January 9, 2008 at 8:30 pm
John smith
so the question is a trick one?
”Find all ordered pairs of prime numbers ……”
because there never was more than 1 pair.
January 9, 2008 at 9:05 pm
Vishal
Well, there are actually two pairs, viz. (2,3) and (3,2), but, yeah, I see what you are saying. On the surface, it does look like a trick question, but then the case where you have to prove that 2^q + q^2 cannot be prime (if q is prime) isn’t that trivial. One still has to present a few arguments (as I did in the solution I wrote) before one can close the proof.
January 15, 2008 at 4:43 pm
John smith
This is one way to solve it. What other ways might there be?
January 15, 2008 at 7:52 pm
Vishal
Perhaps, there are other elementary ways of solving this problem. But, I am not sure how different they will be from the one I presented.