On the Diophantine equation \(3^{2n} - 2 \cdot 3^m+ 1 = k^2\) (Q2889960)
From MaRDI portal
 | This is the item page for this Wikibase entity, intended for internal use and editing purposes. Please use this page instead for the normal view: On the Diophantine equation \(3^{2n} - 2 \cdot 3^m+ 1 = k^2\) |
scientific article; zbMATH DE number 6044085
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | On the Diophantine equation \(3^{2n} - 2 \cdot 3^m+ 1 = k^2\) |
scientific article; zbMATH DE number 6044085 |
Statements
8 June 2012
0 references
exponential Diophantine equations
0 references
Ramanujan-Nagell equation
0 references
linear forms in logarithms
0 references
0.9582142
0 references
0.94918895
0 references
0.9387475
0 references
On the Diophantine equation \(3^{2n} - 2 \cdot 3^m+ 1 = k^2\) (English)
0 references
In this paper the authors completely solve the Diophantine equation \(3^{2n} - 2 \cdot 3^m+ 1 = k^2\). They prove that the only solution is \((n,m,k)=(n,n,3^n-1)\). To obtain the result they show that the Diophantine equation \(x^2+C=3^{2n}\) has at most one solution in positive integers \(x\) and \(n\). The proof of this latter result is based on the identity \((3^n-x)(3^n+x)=C\).
0 references
