# Equation

• Oct 16th 2009, 10:51 AM
streethot
Equation
which points satisfies the problem below?

$\displaystyle XYZ+XZ+XY-YZ-11X-3Y-Z+9=0$
• Oct 16th 2009, 11:11 AM
Mush
Quote:

Originally Posted by streethot
which points satisfies the problem below?

$\displaystyle XYZ+XZ+XY-YZ-11X-3Y-Z+9=0$

x = 0
y = 0
z = 9
• Oct 16th 2009, 11:16 AM
streethot
you have gave a particular solution... now, you have to prove that this is the only solution
• Oct 16th 2009, 12:34 PM
Mush
Quote:

Originally Posted by streethot
you have gave a particular solution... now, you have to prove that this is the only solution

Lol! I'm not here to do your homework for you.

You merely asked for a point that satisfied the equation, and I gave you one.

If you want help with proving or otherwise that there's only one unique solution, then that's a different equation.
• Oct 16th 2009, 01:39 PM
Media_Man
Ah, how the road to mathematical truth is paved in blood...

streethot:

There are more solutions than you should care to count. A simple calculator program can tell you that...

1,-1,-1
1,-1,0
1,-1,1
-1,2,2
0,2,1
1,-1,-2
1,-1,2
-1,3,1
0,1,3
0,3,0

As you know, Diophantine equations are notoriously difficult to solve. There may be an infinite number of solutions readily found by number-crunching.

What exactly is it that you want to know about this equation? A proof of whether or not there is an infinitude of points? A formula generating a set of solutions?
• Oct 16th 2009, 01:51 PM
streethot
Thank you Media_Man. I was thinking that could have a way to solve this problem by the same method used to solve this:

$\displaystyle xyz+xy+yz+zx+x+y+z=243$

A classmate proposed this problem to me, i will investigate. Anyway, thanks for help me
• Oct 16th 2009, 08:56 PM
Bingk
Here's a small part:

If you rearrange the equation you'll get the following:

$\displaystyle Z=\frac{11X+3Y-XY-9}{(X-1)(Y+1)}$

When X=1, and Y=-1 we get $\displaystyle Z=\frac{0}{0}$, in other words $\displaystyle 0 \cdot Z = 0$

So, any Z will work for X=1 and Y=-1

Some other brute force stuff:
When you set one variable to equal zero, there's only a finite number of solutions that will work (tried for X = 0 and Z = 0, Y = 0 will probably behave the same way)

I don't think there's one general formula for the solution though, because of the (1,-1,Z) case ...
• Oct 18th 2009, 08:42 PM
streethot
Now i know what is the real question.

Is for $\displaystyle x,y,z\in\mathbb{N}^*$
• Oct 19th 2009, 02:29 AM
Mush
Quote:

Originally Posted by streethot
Now i know what is the real question.

Is for $\displaystyle x,y,z\in\mathbb{N}^*$

Indeed, it is a diophantine equation.
• Oct 19th 2009, 05:52 AM
streethot
After much work, i got it! i will post with more calm after because the solution is a little long.