Todd and Vishal’s blog

Topological Musings

# Tag Archive

You are currently browsing the tag archive for the ‘theory’ tag.

## Another elementary number theory problem

November 29, 2007 in Problem Corner | Tags: andreescu, elementary, mathematical, number, reflections, theory, titu | by Vishal | 11 comments

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.

## p^q + q^p is prime

November 28, 2007 in Problem Corner | Tags: elementary, Invariant, number, Oxford, prime, theory | by Vishal | 7 comments

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.

• 221,460 hits
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 33, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8980251550674438, "perplexity_flag": "middle"}