Solving the Diophantine Equation (x^3 - 2x - 1 2^n) for Natural Numbers

Diophantine equations are polynomial equations with integer coefficients, for which we seek integer solutions. The equation (x^3 - 2x - 1 2^n) is a particular example that we will explore in detail. This article delves into the methods and reasoning behind solving such equations using modular arithmetic and properties of quadratic residues.

Introduction to the Problem

The problem at hand involves finding all pairs of natural numbers ((n, x)) that satisfy the equation:

[x^3 - 2x - 1 2^n]

Initial Conditions for (n 1) and (n 2)

For (n 1), the equation becomes:

[x^3 - 2x - 1 0]

Using the Rational Root Theorem, we can check that this equation has no integer solutions.

For (n 2), the equation is:

[x^3 - 2x - 3 0]

By direct substitution, we find that (x 1) is the only positive integer solution.

General Case for (n geq 3)

For (n geq 3), we will analyze the equation modulo 8 and modulo 3, and use properties of quadratic residues to deduce the solution:

Modulo 8 Analysis

Reducing both sides of the equation modulo 8, we get:

[x^3 - 2x - 1 equiv 0 pmod{8}]

This congruence is true if and only if (x equiv 5 pmod{8}).

Modulo 3 Analysis

Reducing both sides modulo 3 and using Fermat's Little Theorem (2^2 equiv 1 pmod{3}), we get:

[2^n equiv x^3 - 2x - 1 equiv x - 2x - 1 equiv 1 pmod{3}]

This is only true if (n) is an even number.

Constructing the Solution for (n geq 3)

Let (n 2k) for some (k in mathbb{N}). Adding 2 to both sides of the original equation, we obtain:

[2^{2k} - 2 x cdot x^2 - x - 3]

Given (x equiv 5 pmod{8}), we know that (x^2 - x - 3 equiv 7 pmod{8}). Therefore, there exists a prime (p equiv 5 pmod{8}) or (p equiv 7 pmod{8}) that divides (x^2 - x - 3).

If such a prime (p) does not exist, then (x^2 - x - 3 equiv 1 pmod{8}) or (x^2 - x - 3 equiv 3 pmod{8}).

Since (x^2 - x - 3) divides (2^{2k} - 2), we must have (p | 2^{2k} - 2) or equivalently (2^{k^2} equiv -2 pmod{p}).

However, (-2) is a quadratic residue modulo (p) if and only if (p equiv 1 pmod{8}) or (p equiv 3 pmod{8}), which contradicts our requirement for (p equiv 5 pmod{8}) or (p equiv 7 pmod{8}).

Conclusion and Verification

The above analysis shows that the only solution in positive integers is:

[n 2, x 1]

Therefore, the only pair of natural numbers ((n, x)) that satisfies the equation (x^3 - 2x - 1 2^n) is:

[n 2, x 1]

boxed{n 2, x 1}

References and Further Reading

This solution was obtained using elementary number theory techniques such as modular arithmetic and properties of quadratic residues.

For further reading, consider the following references:

Fred Richman, An Elementary Approach to Quadratic Residues and Quadratic Reciprocity John Derbyshire, Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics