# Thread: Mathematical teaser equation

1. ## Mathematical teaser equation

Hello everyone!

I hope everyone has had a relaxing summer.

Anyway, I have ventured again into a realm of math that is far beyond my abilities, in other words I am trying to solve some mathematical problems from the book Challenging Mathematical Teasers by J.A.H. Hunter

The word problem I am currently solving leads to the following two equations

$\displaystyle x^2-y^2=z^3$

and
$\displaystyle \frac{x}{y}=\frac{y}{z}$

by substiuting we will have $\displaystyle x^2-xz=z^3$

This is about as far as I got, but then the solutions says something that I can't quite understand.

This equation is fully satisfied by: $\displaystyle x-z=mz$, and x=$\displaystyle \frac{z^2}{m}$, where m is any rational number.

Could someone pleade exmplain the above thing to me? Does the author of this book use some systematic approach to solving these kind of equations with 3 unknowns or is it just the mathematician's creativity?

All responses are appreciated.

2. Originally Posted by Coach
Hello everyone!

I hope everyone has had a relaxing summer.

Anyway, I have ventured again into a realm of math that is far beyond my abilities, in other words I am trying to solve some mathematical problems from the book Challenging Mathematical Teasers by J.A.H. Hunter

The word problem I am currently solving leads to the following two equations

$\displaystyle x^2-y^2=z^3$

and
$\displaystyle \frac{x}{y}=\frac{y}{z}$

by substiuting we will have $\displaystyle x^2-xz=z^3$

This is about as far as I got, but then the solutions says something that I can't quite understand.

This equation is fully satisfied by: $\displaystyle x-z=mz$, and x=$\displaystyle \frac{z^2}{m}$, where m is any rational numberd.
This is just a change of variable, what ever the value of $\displaystyle x$ and $\displaystyle z$ we can write $\displaystyle x-z=mz$ for some $\displaystyle m$, then because $\displaystyle x(x-z)=z^3$ we have of necessity that $\displaystyle x=z^2/m$. The advantage is now that by equating the two expressions for $\displaystyle x$ you have a quadratic relating $\displaystyle m$ and $\displaystyle z$, so you can solve this for $\displaystyle m$ in terms of $\displaystyle z$, and so the solutions can be written in tems of a single parameter $\displaystyle z$.

(note you don't need to assume that $\displaystyle m$ is rational, at least I can't see why its necessary unless we are looking for integer or rational solutions)

RonL

3. Coach you've got the equatin
x^2 - xz = z^3
or

x(x-z) = z^3
x-z = z^3/x
now put x= z^2/m

x-z = zm...
so you got it.

4. May i ask a question about "Depreciation"?

5. Originally Posted by Lucy.Gray
May i ask a question about "Depreciation"?
of course! you are more than welcome to... in a new thread