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This one, by Dr. Titu Andreescu (of USAMO fame), is elementary in the sense that the solution to the problem doesn’t require anything more than arguments involving parity and congruences. I have the solution with me but I won’t post it on my blog until Jan 19, 2008, which is when the deadline for submission is. By the way, the problem (in the senior section) is from the 6^{th} issue of Mathematical Reflections, 2007.

Problem: Find the least odd positive integer n such that for each prime \displaystyle p, \, \frac{n^2-1}{4} + np^4 + p^8 is divisible by at least four (distinct) primes.

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 (p,q) such that p^q + q^p is also a prime.

Solution: Let E = p^q+q^p. First, note that if (p,q) is a solution, then so is (q,p). Now, p and q can’t be both even or both odd, else E will be even. Without loss of generality, assume p = 2 and q some odd prime. So, E = 2^q + q^2. There are two cases to consider.

Case 1: q = 3.

This yields E = 2^3 + 3^2 = 17, which is prime. So, (2,3) and, hence (3,2) are solutions.

Case 2: q > 3.

There are two sub-cases to consider.

1^{\circ}: q = 3k+1, where k is some even integer. Then, we have E = 2^{3k+1} + (3k+1)^2 \equiv (-1)^k(-1) + 1 \equiv -1 + 1 \equiv 0 \pmod 3. Hence, 3 \mid E; so, E can’t be prime.

2^{\circ}: q = 3k+2, where k is some odd integer. Then we have E = 2^{3k+2} + (3k+2)^2 \equiv (-1)^k(1) + 1 \equiv -1 + 1 \equiv 0 \pmod 3. Hence, 3 \mid E; so, again, E can’t be prime.

As we have exhausted all possible cases, we conclude (2,3) and (3,2) are the only possible solutions.

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