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 $$x^2+y^2=z^2$$.

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