Possibility of a Separate Forum for NT & Abstract Algebra

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SUMMARY

The discussion emphasizes the need for separate forums dedicated to Number Theory (NT) and Abstract Algebra. Participants explore the mathematical proof demonstrating that there are infinitely many integers \( n \) such that both \( 6n + 1 \) and \( 6n - 1 \) are composite. The proof utilizes specific integer forms, such as \( n = 35x - 1 \) and \( n = 36k^3 \), to illustrate the composite nature of these expressions. The conversation also highlights the absence of LaTeX formatting tools for mathematical expressions in the forum.

PREREQUISITES
  • Understanding of Number Theory concepts
  • Familiarity with Abstract Algebra principles
  • Basic knowledge of composite numbers
  • Experience with mathematical proof techniques
NEXT STEPS
  • Research advanced Number Theory proofs involving composite numbers
  • Explore Abstract Algebra topics related to integer forms and their properties
  • Learn about mathematical proof strategies in Number Theory
  • Investigate the use of LaTeX for formatting mathematical expressions
USEFUL FOR

Mathematicians, students of Number Theory and Abstract Algebra, and anyone interested in exploring the complexities of composite numbers and mathematical proofs.

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1. There should be a separate (sub)forum for NT. ... and one for abstract algebra, for that matter!

2. Show that there are infinitely many n such that both 6n + 1 and 6n - 1 are composite. Without CRT, if possible.

My work... let n = 6^{2k}.
Then 6n \pm 1 = 6^{2k + 1} \pm 1...
Hmm. Having a hard timing finding the LaTexification button from my iPhone...
To be continued.
 
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Let n = 35x - 1, for any positive integer x.

6n + 1 = 6(35x - 1) + 1 = 210x - 5 = 5(42x - 1)
6n - 1 = 6(35x - 1) - 1 = 210x - 7 = 7(30x - 1)

QED

PS: there is no LaTeX yet
 
Bacterius said:
Let n = 35x - 1, for any positive integer x.

6n + 1 = 6(35x - 1) + 1 = 210x - 5 = 5(42x - 1)
6n - 1 = 6(35x - 1) - 1 = 210x - 7 = 7(30x - 1)

QED

PS: there is no LaTeX yet

Slick.
I also like n = 36k^3.
Then 6n \pm 1 is a sum/difference of cubes...
 

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