# Prove that if x,y, and z are integers and

1. Oct 11, 2014

### Bashyboy

1. The problem statement, all variables and given/known data
Prove that if x,y, and z are integers and xyz=1, then x=y=z=1 or two equal -1 and the other is 1.

2. Relevant equations

3. The attempt at a solution
Clearly, if I plug in 1 for each variable, or -1 in for two variables and 1 for the remaining variable, then the equation is satisfied. But how do I know these are the only solutions? This is precisely what I am trying to show. How do I sufficiently demonstrate this?

I thought of writing the equation as $z = \frac{1}{xy}$. But how do I show that, for any choice of integers x and y, besides 1 and -1, z will never be an integer, but a rational number?

2. Oct 11, 2014

### haruspex

What prime factors can x have?

3. Oct 11, 2014

### Bashyboy

I am not sure; I imagine that it would probably be dependent upon certain circumstances, such as the present one, that xyz=1. Is that right?

Last edited: Oct 11, 2014
4. Oct 11, 2014

### haruspex

If x has a factor p, what does that equation tell you?

5. Oct 11, 2014

### Bashyboy

That you should be able to divide both sides by p....? I am not certain.

6. Oct 11, 2014

### HallsofIvy

Staff Emeritus
If x> 1 then xyz> yz so yz must be less than 1.
If x< -1 then xyz< yz so yz must be larger than -1.

Of course, none of x, y, or z can be 0.

7. Oct 11, 2014

### Bashyboy

Would there be a way to show that $\frac{1}{xy}$ is always a rational number, for integers x and y?

8. Oct 11, 2014

### haruspex

That's not exactly right since yz could be negative.
As long as x and y are nonzero integers, it clearly is by definition. But how is that useful here? Stick with the factors hint. If a*b = c and p is a factor of a, what can you say about p and c?

9. Oct 11, 2014

### Bashyboy

Well, I was thinking that this would show that there are no other integer solutions, besides the ones we already mentioned.

10. Oct 11, 2014

### haruspex

Proving a number is rational does not mean it is not an integer.

11. Oct 11, 2014

### Bashyboy

Are not integers those rational numbers whose denominator is one? Well, if x and y are not one, then the denominator of 1/xy would never be one, right?

12. Oct 11, 2014

### Bashyboy

To make this a little more precise, I would have to show that the only solutions to $xy = 1$ are 1 and -1. Here is my proof

$xy = 1$

$x^2 y = x$

$x^2 y - x = 0$

$x(y-1) = 0$

This implies that x=0 or y-1 = 0. But x cannot be 0, as this would not satisfy the equation. Hence, it must be true that $y-1 = 0 \implies y=1$. If y=1, then $x \cdot 1 = 1 \implies x =1$. Notice, however, that I do not get the negative one solution. What is wrong with my proof?

13. Oct 11, 2014

### vela

Staff Emeritus
$x^2 y - x \ne x(y-1)$

14. Oct 11, 2014

### Bashyboy

Ah, yes. I see. So, proving xy=1 is, I imagine, similar to proving that xyz=1. Now, supposing I could prove that xy= 1 implies x=y=1 or x=y=-1, would what I said in this post

still be valid?

15. Oct 11, 2014

### haruspex

Yes, if you add that step it works, but it is rather a long way round. Do try to answer my question about factors.

16. Oct 11, 2014

### Bashyboy

Would it be that p is a factor of c?

17. Oct 11, 2014

### haruspex

Yes. So if c=1 then...?

18. Oct 11, 2014

### Bashyboy

p would have to be one.

19. Oct 12, 2014

### haruspex

If positive, yes, but it could also be...?

20. Oct 12, 2014

### Bashyboy

Or negative one.