<p>Let <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_649_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="61" /> </InlineMediaObject> <EquationSource Format="TEX">\( \{L_n\}_{n\ge 0} \)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mrow> <mo stretchy="false">{</mo> <msub> <mi>L</mi> <mi>n</mi> </msub> <mo stretchy="false">}</mo> </mrow> <mrow> <mi>n</mi> <mo>≥</mo> <mn>0</mn> </mrow> </msub> </math></EquationSource> </InlineEquation> be the sequence of Lucas numbers. In this paper, we look at the exponential Diophantine equation <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_649_Article_IEq2.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="102" /> </InlineMediaObject> <EquationSource Format="TEX">\(L_n-2^x3^y=c\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>L</mi> <mi>n</mi> </msub> <mo>-</mo> <msup> <mn>2</mn> <mi>x</mi> </msup> <msup> <mn>3</mn> <mi>y</mi> </msup> <mo>=</mo> <mi>c</mi> </mrow> </math></EquationSource> </InlineEquation>, for <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_649_Article_IEq3.gif" Format="GIF" Height="18" Rendition="HTML" Resolution="72" Type="Linedraw" Width="92" /> </InlineMediaObject> <EquationSource Format="TEX">\(n,x,y\in \mathbb {Z}_{\ge 0}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>,</mo> <mi>x</mi> <mo>,</mo> <mi>y</mi> <mo>∈</mo> <msub> <mi mathvariant="double-struck">Z</mi> <mrow> <mo>≥</mo> <mn>0</mn> </mrow> </msub> </mrow> </math></EquationSource> </InlineEquation>. We treat the cases <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_649_Article_IEq4.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="55" /> </InlineMediaObject> <EquationSource Format="TEX">\(c\in -\mathbb {N}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>c</mi> <mo>∈</mo> <mo>-</mo> <mi mathvariant="double-struck">N</mi> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_649_Article_IEq5.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="39" /> </InlineMediaObject> <EquationSource Format="TEX">\(c=0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>c</mi> <mo>=</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_649_Article_IEq6.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">\(c\in \mathbb {N}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>c</mi> <mo>∈</mo> <mi mathvariant="double-struck">N</mi> </mrow> </math></EquationSource> </InlineEquation> independently. In the cases that <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_649_Article_IEq6.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">\(c\in \mathbb {N}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>c</mi> <mo>∈</mo> <mi mathvariant="double-struck">N</mi> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10998_2025_649_Article_IEq4.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="55" /> </InlineMediaObject> <EquationSource Format="TEX">\(c\in -\mathbb {N}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>c</mi> <mo>∈</mo> <mo>-</mo> <mi mathvariant="double-struck">N</mi> </mrow> </math></EquationSource> </InlineEquation>, we find all integers <i>c</i> such that the Diophantine equation has at least three solutions. These cases are treated independently, since we employ quite different techniques in proving the two cases.</p>

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On a problem of Pillai involving S-units and Lucas numbers

  • Herbert Batte,
  • Mahadi Ddamulira,
  • Juma Kasozi,
  • Florian Luca

摘要

Let \( \{L_n\}_{n\ge 0} \) { L n } n 0 be the sequence of Lucas numbers. In this paper, we look at the exponential Diophantine equation \(L_n-2^x3^y=c\) L n - 2 x 3 y = c , for \(n,x,y\in \mathbb {Z}_{\ge 0}\) n , x , y Z 0 . We treat the cases \(c\in -\mathbb {N}\) c - N , \(c=0\) c = 0 and \(c\in \mathbb {N}\) c N independently. In the cases that \(c\in \mathbb {N}\) c N and \(c\in -\mathbb {N}\) c - N , we find all integers c such that the Diophantine equation has at least three solutions. These cases are treated independently, since we employ quite different techniques in proving the two cases.