5.3: Non-Linear Diophantine Equations

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.