Factoring x^3+y^3+z^3: Conditions & Solutions

  • Thread starter Thread starter JNeutron2186
  • Start date Start date
  • Tags Tags
    Factoring
JNeutron2186
Messages
7
Reaction score
0
Hey everyone first time poster here, I need help with some factoring of cubes. I know this might tie closely to Diophantine equations but here goes.

Under what condition is the expression x^3+y^3+z^3 factorable? Where x,y,z are positive whole numbers.
 
Physics news on Phys.org
well it seems to factor if x=1, y=2, z=3, and if x=y=z ≥ 2, and more generally if gcd(x,y,z) > 1.
 
My bad for not stating before. Let's assume the gcd(x,y,z) =1 and gcd(x,y)>1
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
Back
Top