<p>Let <InlineEquation ID="IEq5"> <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 Lucas sequence defined by <InlineEquation ID="IEq6"> <EquationSource Format="TEX">\(L_{n + 2} = L_{n + 1} + L_n\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>L</mi> <mrow> <mi>n</mi> <mo>+</mo> <mn>2</mn> </mrow> </msub> <mo>=</mo> <msub> <mi>L</mi> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mi>L</mi> <mi>n</mi> </msub> </mrow> </math></EquationSource> </InlineEquation> for all <InlineEquation ID="IEq7"> <EquationSource Format="TEX">\(n \ge 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>≥</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>, with initial conditions <InlineEquation ID="IEq8"> <EquationSource Format="TEX">\(L_0 = 2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>L</mi> <mn>0</mn> </msub> <mo>=</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq9"> <EquationSource Format="TEX">\(L_1 = 1\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>L</mi> <mn>1</mn> </msub> <mo>=</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>. In this paper, we find all solutions of the Diophantine equations <InlineEquation ID="IEq10"> <EquationSource Format="TEX">\(L_{n} - L_{m} = 5 \cdot 2^a\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>L</mi> <mi>n</mi> </msub> <mo>-</mo> <msub> <mi>L</mi> <mi>m</mi> </msub> <mo>=</mo> <mn>5</mn> <mo>·</mo> <msup> <mn>2</mn> <mi>a</mi> </msup> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq11"> <EquationSource Format="TEX">\(L_{n} - L_{m} = 2 \cdot 5^a\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>L</mi> <mi>n</mi> </msub> <mo>-</mo> <msub> <mi>L</mi> <mi>m</mi> </msub> <mo>=</mo> <mn>2</mn> <mo>·</mo> <msup> <mn>5</mn> <mi>a</mi> </msup> </mrow> </math></EquationSource> </InlineEquation>, where the parameters <i>n</i>,&#xa0;<i>m</i> and <i>a</i> are nonnegative integers such that <InlineEquation ID="IEq12"> <EquationSource Format="TEX">\(n &gt; m\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>&gt;</mo> <mi>m</mi> </mrow> </math></EquationSource> </InlineEquation>. As a corollary, we also determine all solutions of the equation <InlineEquation ID="IEq13"> <EquationSource Format="TEX">\(L_{n} - L_{m} = 10^a\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>L</mi> <mi>n</mi> </msub> <mo>-</mo> <msub> <mi>L</mi> <mi>m</mi> </msub> <mo>=</mo> <msup> <mn>10</mn> <mi>a</mi> </msup> </mrow> </math></EquationSource> </InlineEquation> in nonnegative integers (<i>n</i>,&#xa0;<i>m</i>,&#xa0;<i>a</i>) with <InlineEquation ID="IEq14"> <EquationSource Format="TEX">\(n &gt; m\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>&gt;</mo> <mi>m</mi> </mrow> </math></EquationSource> </InlineEquation>. In order to proof our theorems, we use Baker’s method and various properties of Lucas numbers, along with the theory of lower bounds for linear forms in logarithms of algebraic numbers due to Matveev and Dujella-Pethő version of the reduction method of Baker-Davenport in Diophantine approximation.</p>

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On solutions of the Diophantine equations \(L_{n} - L_{m} = 5 \cdot 2^{a}\) and \(L_{n} - L_{m} = 2 \cdot 5^{a}\)

  • Anouar Gaha,
  • Soufiane Mezroui

摘要

Let \((L_n)_{n \ge 0}\) ( L n ) n 0 be the Lucas sequence defined by \(L_{n + 2} = L_{n + 1} + L_n\) L n + 2 = L n + 1 + L n for all \(n \ge 0\) n 0 , with initial conditions \(L_0 = 2\) L 0 = 2 and \(L_1 = 1\) L 1 = 1 . In this paper, we find all solutions of the Diophantine equations \(L_{n} - L_{m} = 5 \cdot 2^a\) L n - L m = 5 · 2 a and \(L_{n} - L_{m} = 2 \cdot 5^a\) L n - L m = 2 · 5 a , where the parameters nm and a are nonnegative integers such that \(n > m\) n > m . As a corollary, we also determine all solutions of the equation \(L_{n} - L_{m} = 10^a\) L n - L m = 10 a in nonnegative integers (nma) with \(n > m\) n > m . In order to proof our theorems, we use Baker’s method and various properties of Lucas numbers, along with the theory of lower bounds for linear forms in logarithms of algebraic numbers due to Matveev and Dujella-Pethő version of the reduction method of Baker-Davenport in Diophantine approximation.