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5.3: Non-Linear Diophantine Equations

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    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 \(x^2+y^2=z^2\).


    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 \(x^2+y^2=z^2\).


    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.


    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.


    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 license and was authored, remixed, and/or curated by Pamini Thangarajah.