Prove n^2+2^n Composite if n not 6k+3

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Homework Help Overview

The discussion revolves around proving that \( n^2 + 2^n \) is composite for integers \( n > 1 \) that are not of the form \( 6k + 3 \). The subject area includes number theory and properties of composite numbers.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different cases based on the form of \( n \) (e.g., \( 6k, 6k+1, 6k+2, 6k+4, 6k+5 \)). Some suggest checking divisibility by 2 and 3. Questions arise regarding the validity of the proof when \( n = 6k + 3 \) and how it affects the overall argument.

Discussion Status

There is ongoing exploration of various cases and conditions under which \( n^2 + 2^n \) can be shown to be composite. Some participants have provided insights into modular arithmetic and divisibility, while others are questioning the necessity of excluding the case \( n = 6k + 3 \). The discussion remains active with multiple interpretations being considered.

Contextual Notes

Participants note that the original proof relies on specific forms of \( n \) and that the case \( n = 6k + 3 \) has not been fully addressed, raising questions about its implications for the proof's validity.

  • #31
You can write the even case without a ##c##. If ##n## is even, then ##n=2k\, , \,k>0## and ##n^2+2^n=4k^2+4^k\equiv 0 \mod 2## which is a bit shorter.

There are often situations in mathematics where symmetric cases (like odd/even) are not treated by the same methods. One case could be easy, the other one difficult, or require more sub-cases like ##c=1,5## in our case.
 
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