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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.

This one, by Oleg Golberg, appeared in the 6^{th} issue of Mathematical Reflections (MR) 2007. I don’t have a solution yet, but I think I should be able to solve it sooner or later. If you find a solution, you should send it to MR by Jan 19. Here is the problem anyway.

For all integers k, n \geq 2, prove that \\ \displaystyle \sqrt[n]{1 + \frac{n}{k}} \leq \frac1{n} \log \big( 1+\frac{n}{k-1} \big) + 1.

I became interested in mathematical blogging after visiting Terence Tao’s and Timothy Gower’s blogs on numerous occasions. It seems there is a sizable number of mathematicians disseminating valuable information through their blogs, and I see this as a healthy sign. Such blogs provide a wealth of information to students like me, and dare I say, I learn most of my math from such blogs!

I intend to write about math mostly as an exercise in exposition. I am assuming this will be of great help to me later on. I also will be posting some problems in the “Problem Corner” section every now and then.

Let’s see how this goes. I am hoping my enthusiasm for blogging will not wear off too soon!

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