🔢 mathematics

On Carmichael numbers of the form $2^np^m+1$

The paper proves that for any fixed odd integer $m \ge 5$ , there exist only finitely many Carmichael numbers of the form $2^np^m+1$ where $n$ is a positive integer and $p$ is a prime.

Original authors: Florian Luca

Published 2026-06-30

📖 5 min read🧠 Deep dive

CC BY 4.0

Original authors: Florian Luca

Original paper licensed under CC BY 4.0 (http://creativecommons.org/licenses/by/4.0/). ✨ This is an AI-generated explanation of the paper below. It is not written or endorsed by the authors. For technical accuracy, refer to the original paper. Read full disclaimer

Imagine you are a detective trying to solve a mystery about a very specific type of "imposter" number. In the world of mathematics, there are special numbers called Carmichael numbers. These are tricky because they pretend to be prime numbers (the building blocks of math) when you test them with certain rules, even though they are actually made of smaller prime numbers multiplied together.

The paper you provided is a mathematical investigation led by Florian Luca. The goal was to answer a specific question: If we build these imposter numbers using a very specific recipe, how many of them can exist?

Here is the breakdown of the investigation using simple analogies:

1. The Recipe: The "2npm + 1" Cake

The mathematicians are looking at Carmichael numbers that follow a strict recipe:
$N = 2^n \times p^m + 1$
Think of this like baking a cake where:

$p$ is a specific prime ingredient (like "flour").
$m$ is a fixed amount of that ingredient (like "5 cups").
$n$ is a variable amount of another ingredient (like "sugar") that can change.
$2^n$ is a special multiplier (like "yeast" that doubles the size).

The question is: If we fix the amount of flour ( $m$ ) to be an odd number of at least 5, and we keep changing the sugar ( $n$ ) and the type of flour ( $p$ ), how many of these cakes can actually turn out to be "imposter" Carmichael numbers?

2. The Big Discovery: The "Finite" Conclusion

The paper proves a very strong result: There are only a finite number of these cakes.

In other words, even though you could theoretically keep changing the sugar ( $n$ ) and the flour type ( $p$ ) forever, you will eventually run out of combinations that work. You won't find an infinite supply of these specific imposter numbers. Once you pass a certain point, no more will exist.

3. How They Solved It: The "Sieve" and the "Trap"

The proof is like a multi-stage detective story:

Stage 1: The Size Limit (The Sieve)
First, the authors showed that if such a number exists, the amount of sugar ( $n$ ) cannot be arbitrarily huge. It's bounded by the size of the flour ( $p$ ). This narrows the search field significantly. It's like realizing that if a cake is too big, it will collapse, so the baker can't keep making them infinitely large.
Stage 2: The "Multiplicative Independence" Trap
They looked at the prime factors (the ingredients inside the cake). They proved that for these numbers to work, the ingredients must be "independent" in a specific mathematical way. If they weren't, the math would break down (like a cake falling apart). This forced them to conclude that the "order" of the ingredients (how they cycle) must be a power of 2.
Stage 3: The Polynomial Puzzle (The Final Trap)
This is the most complex part. The authors translated the problem into a language of polynomials (equations with variables like $x$ and $y$ ).
- They imagined that if there were infinitely many of these numbers, it would mean a specific polynomial equation had infinitely many solutions.
- They then used a powerful tool from a different branch of math (the Schmidt's Subspace Theorem, mentioned in the paper) to analyze this equation.
- The Analogy: Imagine trying to fit a square peg into a round hole over and over again. The authors showed that if you assume there are infinite solutions, the "pegs" (the mathematical properties of the numbers) would have to be perfect circles. But when they looked closely, the "pegs" were actually squares.
- The Contradiction: They proved that the mathematical "roots" (the solutions) of these equations would have to be "roots of unity" (numbers that circle back to 1). However, the specific structure of their recipe made this impossible. The math simply couldn't balance if there were infinitely many solutions.

4. The Catch: "Ineffective" Proof

The paper admits a limitation. While they proved that the number of these Carmichael numbers is finite, they cannot tell you exactly where the last one is.

The Analogy: It's like a detective saying, "I know the killer stopped committing crimes after a certain date, but I don't know the exact year, and I can't give you a list of the last few crimes."
The proof relies on deep, abstract theorems that guarantee a limit exists but don't provide a calculator to find that limit.

Summary

Florian Luca's paper is a mathematical proof that says: "If you try to build Carmichael numbers using the formula $2^n p^m + 1$ with a fixed odd $m \ge 5$ , you will eventually run out of valid combinations. There is a hard stop; the list of these numbers is not infinite."

They solved this by showing that assuming an infinite list leads to a mathematical contradiction, much like proving a bridge cannot exist because the laws of physics would break if it did.

Technical Summary: On Carmichael Numbers of the Form $2^n p^m + 1$

Problem Statement
The paper investigates the distribution of Carmichael numbers within a specific arithmetic progression. A Carmichael number $N$ is a composite integer satisfying $a^N \equiv a \pmod N$ for all integers $a$ . Such numbers are necessarily odd, squarefree, and satisfy the condition that $p-1 \mid N-1$ for every prime factor $p$ of $N$ . Consequently, every Carmichael number can be written as $N = 2^n k + 1$ for some positive integer $n$ and odd integer $k$ .

The set $K$ is defined as the collection of odd integers $k$ such that $2^n k + 1$ is a Carmichael number for some $n$ . Previous research has established that $K$ contains no primes, no squares of primes, and only the cube $3^3$ . This paper focuses on the case where $k = p^m$ , specifically addressing the following question: For a fixed odd integer $m \ge 5$ , are there infinitely many primes $p$ such that $p^m \in K$ ?

Methodology
The proof proceeds in two main stages, combining elementary number theory with deep results from Diophantine geometry.

Elementary Bounds and Reduction:
The authors first establish an upper bound for the exponent $n$ in terms of $p$ and $m$ . Building on results from [3], they show that if $N = 2^n p^m + 1$ is Carmichael, then $n < 25m^4 \log p$ .
Using this bound, the authors analyze the prime factorization of $N$ . They decompose $N$ into factors of the form $q_i = 2^{a_i} p^{b_i} + 1$ . A critical step involves analyzing the multiplicative dependence of the terms $2^n p^m$ and $2^{a_i} p^{b_i}$ . The authors prove that for sufficiently large $p$ (specifically $p > 500m^5 \log m$ ), the order of $2$ modulo any prime factor $q_i$ of $N$ must be a power of 2. This restricts the structure of the prime factors significantly.
Application of the Subspace Theorem:
Under the assumption that there are infinitely many such primes $p$ , the authors derive a polynomial identity. By fixing the structural parameters (the number of factors $k$ , the exponents $a_i, b_i$ , and the odd parts of $n$ and $a_i$ ), they construct a polynomial equation $P(X, Y) = 0$ where $X = 2^{2\beta'}$ and $Y = p^{\beta'}$ .
The existence of infinitely many solutions $(X, Y)$ with $Y$ being a prime power implies that the polynomial $P(X, Y)$ must vanish identically on a specific algebraic curve. The authors invoke a result by Corvaja and Zannier [4], which relies on Schmidt's Subspace Theorem, to analyze the structure of the polynomials involved.
Specifically, they demonstrate that the polynomials derived from the factorization of $N$ must be related to cyclotomic polynomials. By comparing the degrees and roots of these polynomials, they derive a contradiction: the roots of the derivative of the resulting polynomial would have to lie strictly inside the unit circle, which contradicts the fact that the roots are algebraic integers (via Kronecker's theorem).

Key Contributions and Results
The primary result is Theorem 1:

If $m \ge 5$ is odd, then there are only finitely many primes $p$ such that $p^m \in K$ .

In other words, for any fixed odd $m \ge 5$ , there are only finitely many Carmichael numbers of the form $2^n p^m + 1$ .

The proof establishes that while the set of such primes is finite, the proof is ineffective. The authors explicitly state that they cannot provide an explicit upper bound for the largest such prime $p$ depending on $m$ . The finiteness is derived from the non-existence of infinitely many solutions to the constructed Diophantine equation, rather than an explicit enumeration or bound.

Significance and Claims
The paper contributes to the classification of the set $K$ by extending the known restrictions on the form of elements in $K$ . While previous works ruled out primes, squares of primes, and specific cubes, this work generalizes the result to higher odd powers $p^m$ for $m \ge 5$ .

The authors emphasize that the proof is non-effective, relying on the Schmidt's Subspace Theorem via the work of Corvaja and Zannier. Consequently, while the paper confirms the finiteness of such Carmichael numbers for fixed $m$ , it does not offer a method to compute the specific finite set of exceptions. The result stands as a theoretical limitation on the density of Carmichael numbers within this specific family of arithmetic progressions.

1. The Recipe: The "2npm + 1" Cake

2. The Big Discovery: The "Finite" Conclusion

3. How They Solved It: The "Sieve" and the "Trap"

4. The Catch: "Ineffective" Proof

Summary

More like this