For a positive integer \( n \geq 2 \), does there exist positive integer solutions to the following system of equations? \[ \begin{cases} a^n - 2b^n = 1, \\ b^n - 2c^n = 1. \end{cases} \]
Problem
Source: Israel TST p1
Tags: number theory
Zsigmondy
19.12.2024 23:59
My problem
The answer is no.
We can't have $b=1$, as it would mean that $1-2c^n=1$ which means that $c=0$. Therefore $b>1$. Notice that we have $a^n=2b^n+1,c^n=\frac{b^n-1}{2}$, which means that:
$$b^{2n}-a^nc^n=b^{2n}-(2b^n+1)\cdot\left(\frac{b^n-1}{2}\right)=\frac{b^n+1}{2}$$In particular, $b^{2n}>a^nc^n$ which means that $b^2>ac$. Therefore:
$$\frac{b^n+1}{2}=b^{2n}-a^nc^n \geq b^{2n}-\left(b^2-1\right)^n=b^{2n-2}+b^{2n-4} \left(b^2-1\right) + \ldots + \left(b^2-1\right)^{n-1} \geq b^{2n-2} \geq b^n$$Which is a contradiction.
mathinkingbabo
30.01.2025 04:36
Here's the proof: For n ≥ 2, consider the system of equations: a^n - 2b^n = 1 (1) b^n - 2c^n = 1 (2) From (1), we get: a^n = 2b^n + 1 From (2), we get: b^n = 2c^n + 1 Substituting the second equation into the first: a^n = 2(2c^n + 1) + 1 a^n = 4c^n + 3 Since n ≥ 2, a^n must be at least a^2, and 4c^n + 3 cannot be a perfect square for positive integers a, b, c. This leads to a contradiction. Therefore, there are no positive integer solutions (a, b, c) for this system of equations when n ≥ 2.