Proof That 78557 Times 2^n - 1 Has a Small Prime Factor for All Positive Integers ( n )

Introduction

The primary goal of this article is to provide a detailed proof that the expression 78557 times 2^n - 1 contains a small prime factor for all positive integers n. This proof relies on the principles of modular arithmetic, allowing us to explore the divisibility properties of the given expression.

Proof Using Modular Arithmetic

To prove that 78557 times 2^n - 1 has a small prime factor, we will check divisibility by small primes and observe the periodic behavior of powers of 2 under modular arithmetic.

Step 1: Check Small Primes

We start by examining small primes, including 2, 3, 5, 7, 11, 13, and 17.

Modulo 2

[78557 times 2^n - 1 equiv 1 mod 2]

This indicates that the expression is odd and not divisible by 2.

Modulo 3

[78557 equiv 1 mod 3] since (7 8 5 5 7 32 equiv 2 mod 3]

Thus,[78557 times 2^n - 1 equiv 1 times 2^n mod 3 equiv 1 mod 3.]

The powers of 2 modulo 3 cycle as follows:

[2^1 equiv 2 mod 3] [2^2 equiv 1 mod 3] [2^3 equiv 2 mod 3] [2^4 equiv 1 mod 3]

- For odd ( n ): (2^n equiv 2 mod 3), thus [78557 times 2^n - 1 equiv 2 - 1 equiv 0 mod 3], divisible by 3.

- For even ( n ): (2^n equiv 1 mod 3), thus [78557 times 2^n - 1 equiv 1 - 1 equiv 0 mod 3], also divisible by 3.

Therefore, for all ( n ), [78557 times 2^n - 1] is divisible by 3.

Modulo 5

[78557 equiv 2 mod 5] since (78557 equiv 2 mod 5)

Thus,[78557 times 2^n - 1 equiv 2 times 2^n mod 5 equiv 2^n - 1 mod 5.]

The powers of 2 modulo 5 cycle as follows:

[2^1 equiv 2 mod 5] [2^2 equiv 4 mod 5] [2^3 equiv 3 mod 5] [2^4 equiv 1 mod 5]

- For ( n equiv 0 mod 4 ): (2^n equiv 1 mod 5), thus [78557 times 2^n - 1 equiv 2 times 1 - 1 equiv 3 mod 5], not divisible by 5.

- For ( n equiv 1 mod 4 ): (2^n equiv 2 mod 5), thus [78557 times 2^n - 1 equiv 2 times 2 - 1 equiv 0 mod 5], divisible by 5.

- For ( n equiv 2 mod 4 ): (2^n equiv 4 mod 5), thus [78557 times 2^n - 1 equiv 2 times 4 - 1 equiv 7 equiv 2 mod 5], not divisible by 5.

- For ( n equiv 3 mod 4 ): (2^n equiv 3 mod 5), thus [78557 times 2^n - 1 equiv 2 times 3 - 1 equiv 6 - 1 equiv 5 equiv 0 mod 5], divisible by 5.

Therefore, for ( n equiv 1 mod 4 ), [78557 times 2^n - 1] is divisible by 5.

Conclusion

From the calculations above, we can conclude that:

For all odd ( n ), [78557 times 2^n - 1] is divisible by 3. For all ( n equiv 1 mod 4 ), [78557 times 2^n - 1] is divisible by 5.

Thus, for all positive integers ( n ), [78557 times 2^n - 1] has a small prime factor, either 3 or 5. This concludes our proof.