Factoring b^n + 1

The previous post illustrated a technique for finding factors of number of the form bⁿ - 1. This post will look at an analogous, though slightly less general, technique for numbers of the form bⁿ + 1.

There is a theorem that says that if m divides n then b^m + 1 divides bⁿ + 1 if n is odd. To see that the exponent must be odd, let b = 10, n = 4, and m = 2: 10,0001 is not divisible by 101.

Last time we looked at 2²⁰²³ - 1, so this time let's look at J = 2²⁰²³ + 1.

Since 2023 = 7*17*17, we know J is divisible by 2ⁿ + 1 for n = 7, 17, 7*17, and 17*17. Factoring each of these factors tells us that J is divisible by

• 3
• 43691
• 72251
• 79187
• 1077971
• 823679683
• 18360250452977
• 197766803208315851
• 143162553165560959297
• 338858733065598401355195539629373089

(If it seems this post is moving too fast, you might want to go back and read the previous post and then come back.)

Let P be the product of the factors above, and let F = J/P. Now F is a 493-digit number, so we don't expect to be able to get much further unless we're lucky. We can find three factors of F

• 43
• 4382432993
• 1809032819104754041

but then we get stuck. We know there are more factors, but it seems doubtful that I'm going to find any more anytime soon.

***

As in the previous post, we'll conclude with Python code to verify the claims above.

from sympy import isprimedef verify_factors(N, factors, full=True): prod = 1 for f in factors: assert(isprime(f)) prod *= f if full: assert(N == prod) else: assert(N % prod == 0) assert(not isprime(N // prod)) return N // prodfactors = [ 3, 43691, 72251, 79187, 1077971, 823679683, 18360250452977, 197766803208315851, 143162553165560959297, 338858733065598401355195539629373089]J = 2**2023 + 1F = verify_factors(J, factors, False)factors = [43, 4382432993, 1809032819104754041]verify_factors(F, factors, False)print(len(str(F)))

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