You are currently browsing the daily archive for November 28, 2007.

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!

• 221,624 hits