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?

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    $\begingroup$ I think you mean $DX^2$ not $(DX)^2$, right? This would be consistent with the solutions you present. (Great puzzle btw!) $\endgroup$ – Rand al'Thor Sep 26 '20 at 19:30
  • $\begingroup$ I meant $ (DX)^2$. $\endgroup$ – Vassilis Parassidis Sep 26 '20 at 19:47
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    $\begingroup$ But with $(DX)^2$ your proposed solutions are not solutions, since they involve "something times a square, minus a square, equals a fifth power" and not "a square minus a square equals a fifth power". $\endgroup$ – Rand al'Thor Sep 26 '20 at 19:48
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    $\begingroup$ Then can you explain how $11*(641)^2-(2122)^2=7^5$ is a solution of the expression $(DX)^2-Y^2=\pm Z^5$? It seems rather to be a solution of the expression $DX^2-Y^2=\pm Z^5$. $\endgroup$ – Rand al'Thor Sep 26 '20 at 20:16
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    $\begingroup$ But wouldn't it make a lot more sense to write $DX^2-Y^2=Z^5$? Then the numbers $D,X,Y,Z$ are all integers, and the equation looks like the one that you've solved in your examples rather than just "difference of two squares". $\endgroup$ – Rand al'Thor Sep 26 '20 at 22:26

The calculation process

Given $x$ and $y$, calculate


so that


Taking powers, this means


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


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.

  • $\begingroup$ Can you present a couple of numerical examples with your thought process? $\endgroup$ – Vassilis Parassidis Sep 26 '20 at 19:50
  • $\begingroup$ @Vassilis Sure, done. $\endgroup$ – Rand al'Thor Sep 26 '20 at 19:58
  • $\begingroup$ You did a good job Rand. FYI $Y= y^5 + 10xy^3 + 5x^2y$. Now you can find what $DX$ is equal to. $\endgroup$ – Vassilis Parassidis Sep 26 '20 at 20:21
  • $\begingroup$ @Vassilis $X=5y^4+10xy^2+x^2$ and $D=\sqrt{x}$? $\endgroup$ – Rand al'Thor Sep 26 '20 at 20:23
  • $\begingroup$ You are very close.Give it another try and then I will reveal the answer. $\endgroup$ – Vassilis Parassidis Sep 26 '20 at 20:49

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