We classify all solution triples with k-Fibonacci components to the equation <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="9_2025_2845_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="181" /> </InlineMediaObject> <EquationSource Format="TEX">$x^2+y^2+z^2=3xyz+m$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>y</mi> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>z</mi> <mn>2</mn> </msup> <mo>=</mo> <mn>3</mn> <mi>x</mi> <mi>y</mi> <mi>z</mi> <mo>+</mo> <mi>m</mi> </mrow> </math></EquationSource> </InlineEquation>, where m is a positive integer and <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="9_2025_2845_Article_IEq2.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">$k\ge 2$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>k</mi> <mo>≥</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>. As a result, for <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="9_2025_2845_Article_IEq3.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="46" /> </InlineMediaObject> <EquationSource Format="TEX">$m=8$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>m</mi> <mo>=</mo> <mn>8</mn> </mrow> </math></EquationSource> </InlineEquation>, we have the Markoff triples with Pell components <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="9_2025_2845_Article_IEq4.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="189" /> </InlineMediaObject> <EquationSource Format="TEX">$(F_2(2), F_2(2n), F_2(2n+2))$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <msub> <mi>F</mi> <mn>2</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mn>2</mn> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <msub> <mi>F</mi> <mn>2</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mn>2</mn> <mi>n</mi> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <msub> <mi>F</mi> <mn>2</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mn>2</mn> <mi>n</mi> <mo>+</mo> <mn>2</mn> <mo stretchy="false">)</mo> </mrow> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation>, for <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="9_2025_2845_Article_IEq5.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="43" /> </InlineMediaObject> <EquationSource Format="TEX">$n\ge 1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>≥</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>. For all other m there exists at most one such ordered triple, except when <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="9_2025_2845_Article_IEq6.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="41" /> </InlineMediaObject> <EquationSource Format="TEX">$k=3$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>k</mi> <mo>=</mo> <mn>3</mn> </mrow> </math></EquationSource> </InlineEquation>, a is odd, b is even and <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="9_2025_2845_Article_IEq7.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="69" /> </InlineMediaObject> <EquationSource Format="TEX">$b\ge a+3$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>b</mi> <mo>≥</mo> <mi>a</mi> <mo>+</mo> <mn>3</mn> </mrow> </math></EquationSource> </InlineEquation>, where <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="9_2025_2845_Article_IEq8.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="169" /> </InlineMediaObject> <EquationSource Format="TEX">$(F_3(a),F_3(b),F_3(a+b))$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <msub> <mi>F</mi> <mn>3</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>a</mi> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <msub> <mi>F</mi> <mn>3</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>b</mi> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <msub> <mi>F</mi> <mn>3</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>a</mi> <mo>+</mo> <mi>b</mi> <mo stretchy="false">)</mo> </mrow> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="9_2025_2845_Article_IEq9.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="226" /> </InlineMediaObject> <EquationSource Format="TEX">$(F_3(a+1),F_3(b-1),F_3(a+b))$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <msub> <mi>F</mi> <mn>3</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>a</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <msub> <mi>F</mi> <mn>3</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>b</mi> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <msub> <mi>F</mi> <mn>3</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>a</mi> <mo>+</mo> <mi>b</mi> <mo stretchy="false">)</mo> </mrow> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> share the same m.

Markoff m-Triples with k-Fibonacci Components

期刊论文
发表日期：

2025年4月18日

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D. Alfaya,
L. A. Calvo,
A. Martínez de Guinea,
J. Rodrigo,
A. Srinivasan

摘要

We classify all solution triples with k-Fibonacci components to the equation $x^2+y^2+z^2=3xyz+m$ $x^{2} + y^{2} + z^{2} = 3 x y z + m$ , where m is a positive integer and $k\ge 2$ $k \geq 2$ . As a result, for $m=8$ $m = 8$ , we have the Markoff triples with Pell components $(F_2(2), F_2(2n), F_2(2n+2))$ $(F_{2} (2), F_{2} (2 n), F_{2} (2 n + 2))$ , for $n\ge 1$ $n \geq 1$ . For all other m there exists at most one such ordered triple, except when $k=3$ $k = 3$ , a is odd, b is even and $b\ge a+3$ $b \geq a + 3$ , where $(F_3(a),F_3(b),F_3(a+b))$ $(F_{3} (a), F_{3} (b), F_{3} (a + b))$ and $(F_3(a+1),F_3(b-1),F_3(a+b))$ $(F_{3} (a + 1), F_{3} (b - 1), F_{3} (a + b))$ share the same m.