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 smithbut 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

AnonymousNote 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

VishalThe above was my comment actually.

January 9, 2008 at 8:30 pm

John smithso 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

VishalWell, 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

thattrivial. 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 smithThis is one way to solve it. What other ways might there be?

January 15, 2008 at 7:52 pm

VishalPerhaps, there are other elementary ways of solving this problem. But, I am not sure how different they will be from the one I presented.