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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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