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

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SUMMARY

The discussion centers on proving that for any integer \( n > 1 \) not of the form \( 6k + 3 \), the expression \( n^2 + 2^n \) is composite. The proof demonstrates that if \( n \) is even, \( n^2 + 2^n \) is divisible by 2, and if \( n \) is odd, it can be shown that \( n^2 + 2^n \) is divisible by 3. The cases considered include \( n = 6k, 6k + 1, 6k + 2, 6k + 4, \) and \( 6k + 5 \), confirming that \( n^2 + 2^n \) is composite for these forms. The case \( n = 6k + 3 \) is excluded, as it leads to a prime result.

PREREQUISITES
  • Understanding of modular arithmetic, particularly modulo 2 and 3.
  • Familiarity with the Division Algorithm and its application in number theory.
  • Basic knowledge of composite and prime numbers.
  • Ability to manipulate and simplify algebraic expressions involving powers.
NEXT STEPS
  • Study the properties of composite numbers and their factors.
  • Learn about modular arithmetic and its applications in number theory.
  • Explore proofs involving divisibility and prime number theorems.
  • Investigate further cases of \( n^2 + 2^n \) for different forms of \( n \).
USEFUL FOR

This discussion is beneficial for mathematicians, number theorists, and students interested in proofs involving composite numbers and modular arithmetic. It is particularly useful for those studying properties of integers in relation to prime factorization.

  • #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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