🔢 mathematics

Sum of consecutive powers as a perfect power

This paper proves that for the equation $x^k + (x+1)^k = y^n$ with $n \geq 3$ and $k \equiv 2 \pmod{4}$ , the only solutions are the trivial cases $x=0, -1$ when $6 \leq k \leq 100$ or when $k$ possesses odd prime factors congruent to $3 \pmod{4}$ , utilizing linear forms in logarithms, the modular method, and Thue equations.

Original authors: Angelos Koutsianas, Nikos Tzanakis

Published 2026-05-19

📖 5 min read🧠 Deep dive

CC BY 4.0

Original authors: Angelos Koutsianas, Nikos Tzanakis

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 very specific, stubborn riddle involving numbers. The riddle is this: Can you find two numbers that sit right next to each other (like 3 and 4, or 100 and 101), raise them both to the same high power, add them together, and get a result that is also a perfect power?

For example, if you take $3^3 + 4^3$ , you get $27 + 64 = 91$ . Is 91 a perfect power (like a square, cube, etc.)? No. The mathematicians in this paper, Angelos Koutsianas and Nikos Tzanakis, spent their time hunting for the rare moments when the answer is yes.

Here is a breakdown of their investigation using simple analogies.

The Main Riddle

The equation they are studying looks like this:
$x^k + (x+1)^k = y^n$

$x$ and $x+1$ : Two neighbors.
$k$ : The "power" you raise them to. The paper focuses on a specific type of power: numbers like 6, 10, 14, 18, etc. (numbers that are 2 more than a multiple of 4).
$y^n$ : The result must be a perfect power (like a perfect cube, fourth power, etc., where $n$ is at least 3).

The "Ghost" Solutions

Before they started, they knew about two "ghost" solutions.

If $x = 0$ , then $0^k + 1^k = 1$ . Since $1$ is a perfect power ( $1^n$ ), this works.
If $x = -1$ , then $(-1)^k + 0^k = 1$ . This also works.

The authors wanted to know: Are there any real solutions where the numbers are actually bigger than 1?

The Detective's Toolkit

To solve this, the authors didn't just guess numbers. They used a "three-pronged" investigation strategy, like a detective using three different types of forensic tools:

The "Logarithm Tape Measure" (Linear Forms in Logarithms):
Imagine you are trying to find a needle in a haystack. You know the needle is somewhere, but the haystack is infinite. This tool helps them cut the haystack down to a manageable size. It proves that if a solution exists, the numbers involved can't be too huge. It sets an upper limit, saying, "If a solution exists, it must be below this specific number."
The "Modular Mirror" (The Modular Method):
This is the most complex tool. Imagine looking at a reflection in a mirror. If you have a specific shape (your equation), it casts a shadow (a mathematical object called an elliptic curve). The authors looked at this shadow and compared it to a library of known shadows (called "newforms").
- If the shadow of their equation matched a shadow in the library, they could check if it was a "fake" match.
- They used this to prove that for many specific powers ( $k$ ), the "shadow" didn't match anything that could produce a real solution. It's like saying, "This fingerprint doesn't belong to any criminal in our database, so this crime couldn't have happened."
The "Puzzle Solver" (Thue Equations):
For the smaller numbers that the other tools couldn't rule out, they turned the problem into a specific type of mathematical puzzle called a Thue equation. These are like Sudoku grids for numbers. They solved these puzzles for smaller cases to prove that no solutions existed there either.

The Big Discovery

After running these tools through a massive computer check (using a standard office computer, not a supercomputer), they found the answer:

For every power $k$ between 6 and 100 (that fits their specific rule), the ONLY solutions are the "ghosts" we already knew: $x = 0$ and $x = -1$ .

In other words, you cannot take two positive integers sitting next to each other, raise them to a power between 6 and 100, add them up, and get a perfect power. The universe simply doesn't allow it in this range.

Why This Matters (In Math Terms)

The paper mentions that this is the first time anyone has successfully solved this specific type of riddle for such large powers ( $k$ ) when there are only two numbers being added.

Usually, when you have two numbers adding up to a third power, it's a very hard problem (related to the famous Fermat's Last Theorem). The authors managed to break the problem down into smaller, solvable pieces. They showed that even though the math is incredibly deep and involves abstract concepts like "curves" and "logarithms," the final result is a simple "No" for all the big numbers they tested.

The "What's Next?"

The paper ends by saying, "We did this for powers up to 100. The same methods should work for powers bigger than 100, but that's a job for another day." They also note that if the power $k$ is an odd number (like 5 or 7), the whole game changes, and the rules they used here no longer apply. That is a much harder mystery for the future.

In short: They proved that for a wide range of high powers, the sum of two consecutive numbers is never a perfect power, unless you start with zero or negative one. The "ghosts" are the only ones who win.

Technical Summary: Sum of Consecutive Powers as a Perfect Power

Problem Statement
This paper investigates the Diophantine equation representing the sum of two consecutive perfect powers equating to a perfect power:
$x^k + (x + 1)^k = y^n, \quad n \geq 3$
The authors specifically focus on the case where the exponent $k$ satisfies $k \equiv 2 \pmod 4$ . While the cases for $k=2, 3, 4$ and various lengths of consecutive sums ( $d$ ) have been addressed in prior literature, the case $d=2$ with large $k$ remains largely unresolved. The authors note that for $d=2$ , the equation is a special case of the generalized Fermat equation of signature $(r, r, p)$ , which typically presents obstructions related to non-singular Frey hyperelliptic curves with complex multiplication (specifically for $x=0, -1$ ). The primary goal is to determine all integer solutions for $6 \leq k \leq 100$ under the condition $k \equiv 2 \pmod 4$ .

Methodology
The proof strategy combines three distinct mathematical techniques: linear forms in logarithms, the modular method, and the resolution of Thue equations. The approach proceeds as follows:

Transformation and Reduction:
The authors transform the original equation by setting $x$ to be an odd integer (via variable substitution $x \to (x-1)/2$ ), leading to the symmetric form:
$\frac{(x-1)^k + (x+1)^k}{2} = 2^{k-1}y^n$
Using Lemma 4.1, they factor the left-hand side as $(x^2+1)f_k(x^2)$ , where $f_k$ is a polynomial of degree $k/2 - 1$ .
Reduction to Lebesgue-Nagell and Thue Equations:
A key contribution is Lemma 4.2, which demonstrates that for $k \geq 6$ and sufficiently large $n$ , any solution $(x, y, n)$ reduces to a finite system of two equations:
$x^2 + 1 = 2 \frac{d_2}{d_1} y_1^n$
$f_k(x^2) = 2^{k-2} \frac{d_1}{d_2} y_2^n$
Here, $d_1, d_2$ are specific divisors of $k/2$ determined by the prime factors of $k$ congruent to $1 \pmod 4$ .
- For small $n$ ( $n \leq 7$ ), the resolution of the first equation is reduced to solving a finite number of Thue equations over the Gaussian integers $\mathbb{Q}(i)$ .
- For the general case, the problem is reduced to solving Lebesgue-Nagell equations of the form $x^2 + 1 = 2d y^n$ .
The Modular Method:
For the range $6 \leq k \leq 100$ and prime $n \geq 11$ , the authors employ the modular method. They associate a Frey-Hellegouarch elliptic curve $E$ to the potential solution. By utilizing modularity and level lowering theorems, they establish an isomorphism between the Galois representation of $E$ and that of a newform $f$ .
- Proposition 3.1 is used to bound $n$ by comparing Fourier coefficients $a_\ell(f)$ and $a_\ell(E)$ for various primes $\ell$ .
- Proposition 6.3 introduces a refined elimination technique using primes $\ell = tn + 1$ to rule out specific newforms that standard coefficient comparison could not eliminate.
Linear Forms in Logarithms:
To handle the upper bound of $n$ , the authors apply Baker's theory of linear forms in logarithms. They derive an explicit upper bound $n_0$ for $n$ (presented in Table 2) by analyzing the linear form $\Lambda = n \log \alpha_1 - \log \alpha_2$ . This bound ensures that for $n > n_0$ , no solutions exist.

Key Results
The paper establishes the following main theorem:

Theorem 1.1: Suppose $k \equiv 2 \pmod 4$ . If $6 \leq k \leq 100$ , the only integer solutions to $x^k + (x+1)^k = y^n$ with $n \geq 3$ occur when $x \in \{-1, 0\}$ .

Additionally, the authors prove Theorem 1.3, which addresses the case where $k \equiv 2 \pmod 4$ and all odd prime factors of $k$ are congruent to $3 \pmod 4$ . In this specific scenario, the solutions are $(x, y, n, k) = (0, 1, n, k)$ , $(119, 13, 4, 2)$ , and $(-120, 13, 4, 2)$ .

The computational verification involved:

Solving Thue equations for $n \leq 7$ using Pari/GP.
Applying the modular method to eliminate newforms for $11 \leq n \leq n_0$ using Sage scripts.
Calculating explicit upper bounds $n_0$ for each $k$ in the range $[6, 100]$ (ranging from $\approx 1.7 \times 10^5$ to $\approx 3 \times 10^7$ ).

Significance and Scope
The authors state that this work represents the first attempt to solve equation (1.1) for $d=2$ with large values of $k$ . The significance lies in successfully overcoming the obstructions associated with genus $>1$ curves (specifically the $x=0, -1$ cases) by reducing the problem to a finite number of solvable equations.

The paper is modest regarding its scope beyond the tested range. The authors note that while their methods are "in principle" applicable to any $k \equiv 2 \pmod 4$ with $k > 100$ , the computational complexity increases significantly. They explicitly state that the case where $k$ is odd presents a "very different" and "very challenging" situation where their key Lemma 4.2 fails, and the case $k=5$ remains an open problem. The paper does not claim to solve the equation for all $k$ , but rather provides a complete resolution for the specific range $6 \leq k \leq 100$ under the stated congruence conditions.