Skip to main content
Mathematics LibreTexts

5.3: Non-Linear Diophantine Equations

  • Page ID
    7599
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    The following are some well-known examples of non-linear Diophantine equations:

    Pythagorean Equation

    Pythagorean Equation

    Equations of the form \(x^2+y^2=z^2\), where \(x,y,z \in \mathbb{Z}\).

    This equation arises out of geometric consideration, as Pythagoras was a geometer in ancient Greece.

    Note that \(3^2+4^2=5^2\) is a solution to the above equation. In this case, \(4+5=3^2\).

    This happens to be the first instance of a pattern.

    Example \(\PageIndex{1}\)

    Let \(x\) be a positive odd integer. If \(x^2\) is a sum of two consecutive positive integers \(y\) and \( z\), then \(x^2+y^2=z^2\).

    Solution

    Let \(x,y,z \in \mathbb{Z}_+\) such that  \(x^2=y+z\) and \(z=y+1\).

    Consider, \(x^2+y^2=(y+z)+y^2=(y+y+1)+y^2=y^2+2y+1=(y+1)^2=z^2.\)

    Note that \(5^2=12+13\) and \(7^2=24+25\).

    Example \(\PageIndex{2}\)

    Let \(x\) be a positive even integer. If \(\dfrac{x^2}{2}\) is a sum two positive integers \(y\) and \( z\) differs by \(2\), then \(x^2+y^2=z^2\).

    Solution

    Note that \(\dfrac{6^2}{2}=18=10+8\) and \(10^2=8^2+6^2\).

    Also, \(\dfrac{8^2}{2}=32=15+17\) and \(17^2=15^2+8^2\).

    Note that these patterns always generate solutions to the Pythagorean equations.  Thus there are infinitely many solutions to the Pythagorean equations.  Note that \(\{x=3,y=4, z=5\}\) and \(\{x=6,y=8,z=10\}\) are solutions to the Pythagorean equations.

    However, these patterns do not generate all solutions.

    Example \(\PageIndex{3}\)

    Not all the solutions to   \(x^2+y^2=z^2\) ,  \(x,y,z \in \mathbb{Z}_+\) can be obtained by doubling a solution.

    Solution

    Note that \(\{x=20,y=21, z=29\}\) can't be obtained from doubling a solution.

    Pellian Equation

    Pellian Equation

    Equations of the form \(x^2-dy^2=1\), where \(x,y \in \mathbb{Z}\), and \(d\) is a positive integer which is not a square of an integer.

    Note that the solutions to \(x^2-2y^2=1\), where \(x,y \in \mathbb{Z}\), give rise to square triangular numbers.

     A square triangular numbers are of the form \(\dfrac{t(t+1)}{2}=s^2\), for some \(t, s \in \mathbb{Z}_+.\)

    Then \(t^2+t=2s^2.\) Now \(4(t^2+t)+1=8s^2+1\).

    Thus \((2t+1)^2=8s^2+1.\)  Let \(x=2t+1\) and \(y=2s\), then \(x^2-2y^2=1.\)

    Example \(\PageIndex{4}\)

    Consider the Pellian equation \(x^2-2y^2=1\).

    1. Find a solution to the Pellian equation, by inspection.
    2. Prove that if \(x=a\) and \(y=b\) is a solution, then \(x=a^2+2b^2\) and \(y=2ab\) is also a solution.

    Solution

    1.\( x=3, y=2\).

    2. Assume that \(a^2-2b^2=1\). Consider \((a^2+2b^2)^2-2(2ab)^2= a^4+4a^2b^2+4b^4-8a^2b^2=a^4-4a^2b^2+4b^4=(a^2-2b^2)^2=1.\) Hence the result.


    This page titled 5.3: Non-Linear Diophantine Equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.