Unusual connections of numbers

Question

Let's have the equation $(DX)^2-Y^2= ± Z^5$ and $x,y$ two positive integers greater than zero. From some facts we can obtain solutions of the above equation by giving integer values at $x,y$. Examples:

• if $x=11$ and $y=2$ then we have the solution $11*(641)^2-(2122)^2=7^5$;

• if $x=5$ and $y=2$ then we have the solution $5*(305)^2-(682)^2=1^5$;

• if $x=7$ and $y=4$ then we have the solution $7*(2449)^2-(6484)^2=(-9)^5$.

The puzzle is:

1. Can you present more such solutions?

2. What is the trick to solving such equations?

Answer 1

The calculation process

Given $x$ and $y$, calculate

$x-y^2=Z$,

so that

$(y-\sqrt{x})(y+\sqrt{x})=-Z$.

Taking powers, this means

$(y-\sqrt{x})^5(y+\sqrt{x})^5=-Z^5$.

Now we know that

$(y+\sqrt{x})^5$ is a number in the form $Y+X\sqrt{x}$ for some integers $X,Y$, and then by conjugation $(y-\sqrt{x})^5=Y-X\sqrt{x}$.

So we have found a solution

$(Y-X\sqrt{x})(Y+X\sqrt{x})=-Z^5$,

or in other words $DX^2-Y^2=Z^5$ with

$D=x$, $X=5y^4+10xy^2+x^2$, $Y=y^5+10xy^3+5x^2y$,

or in the OP's notation $(DX)^2-Y^2=Z^5$ with

$D=\sqrt{x}$, $X=5y^4+10xy^2+x^2$, $Y=y^5+10xy^3+5x^2y$.

An example

Let's start with the relatively simple example $x=2,y=1$ and calculate as follows:

• $Z=1$, $(1-\sqrt{2})(1+\sqrt{2})=-1$,

• $(1+\sqrt{2})^5=1+5(\sqrt{2})+10(2)+10(2\sqrt{2})+5(4)+(4\sqrt{2})=41+29\sqrt{2}$,

• $(41-29\sqrt{2})(41+29\sqrt{2})=(-1)^5$,

and the solution $2*(29)^2-(41)^2=1^5$, which can be checked by calculator.

As another example with an even value of $Z$, let's use $x=3,y=1$ and get:

• $Z=2$, $(1-\sqrt{3})(1+\sqrt{3})=-2$,

• $(1+\sqrt{3})^5=1+5(\sqrt{3})+10(3)+10(3\sqrt{3})+5(9)+(9\sqrt{3})=76+44\sqrt{3}$,

• $(76-44\sqrt{3})(76+44\sqrt{3})=(-2)^5$,

and the solution $3*(44)^2-(76)^2=2^5$, which can be checked by calculator.

My thought process

This whole puzzle seems likely to be something to do with

Pell's equation and the associated theory of algebraic number fields,

indeed the second of the examples given in the OP is actually a solution of this equation. The new feature in this puzzle, compared with the stuff I learned in undergraduate number theory courses, is the involvement of fifth powers. How can we relate fifth powers to

the square-root arithmetic that enables solutions of Pell's equation? Well, if we have $a-b\sqrt{d}$ giving rise to an integer $n$ when multiplied by its conjugate, then $(a-b\sqrt{d})^5$ will similarly give rise to the integer $n^5$.

I then spotted the pattern in the given values that

$11-2^2=7$, $5-2^2=1$, $7-4^2=-9$,

which confirmed for me the direction in which to approach this puzzle.