1. Feb 12, 2004

### MathematicalPhysicist

i think everyone is familiar to this but i wonder if someone has made his research to study it.
what i mean is the namubers that when the result of multiplying them together is the same as adding them:
for example:
2*2=2+2=4 (trivial, i know (-: ).
3*1.5=3+1.5=4.5

my question has someone prooved that the number of these pairs are finite or infinite in the real numbers?

2. Feb 12, 2004

### matt grime

xy=x+y

rearrange and solve for R.

for N (well, this is the number theory thread)

xy-x-y+1=1

(x-1)(y-1)=1

in N x=y=2

3. Feb 12, 2004

### Muzza

Perhaps I'm misunderstanding something... You want real solutions to this equation:

x + y = xy
=>
y = x / (x - 1)

I.e, for any x != 1 we can chose a y so that x + y = xy, namely y = x / (x - 1). Any real x (except for x = 1) will give a real y, so there are an infinite number of such pairs...

A more interesting question would be to consider only integers...

4. Feb 12, 2004

### MathematicalPhysicist

i didnt think about it very much so perhaps from this had risen the misunderstanding.
i guess in the real numbers is really trivial.
so lets say just prooving for integers how do you go around prooving this?
[x,y]=[Z] or [x,x/(x-1)]=[Z]

5. Feb 12, 2004

### Muzza

I have no idea what this means. matt grime provided an answer for what happens if you only consider the natural numbers (i.e, x = y = 2 is the only solution).

Last edited: Feb 12, 2004
6. Feb 12, 2004

### MathematicalPhysicist

z doesnt represent the integers?

7. Feb 12, 2004

### Muzza

Aha, yes it does, but you used the equals sign (and non-curly braces), and I've never seen = used to specify membership in a set ( seems to be more commonly used).

Last edited: Feb 12, 2004
8. Feb 12, 2004

### NateTG

For $$\frac{x}{x-1}$$ to be an integer, $$x-1$$ must divide$$x$$. Since the difference between the two is 1, their greatest common factor is 1, so the only solutions are $$x-1=1$$ and $$x-1=-1$$ (2+2=2*2 and 0+0=0*0).

9. Feb 12, 2004

### matt grime

NateG, you should state you are only working in N there, by saying integer you implying Z. In Z, there are other answers. All we know is that here (x-1) is a unit, hence x-1=-1 or 1, yielding the other integer answer of x=y=0 as well as x=y=2

10. Feb 12, 2004

### NateTG

For $$\frac{x}{x-1}$$ to be an integer |x-1| must divide |x|. The GCF of the two absolute values is 1 (by Euclid's algorithm). Therefore $$|x-1| \leq 1$$ since it divides |x| and obviously divides itself. For $$x \in \mathbb{Z}$$ that leaves three solutions: 0,1, and 2, but 1 leads to division by zero, so the only solutions are x=0 and x=2.

P.S. My apologies for the mixed formatting

Last edited: Feb 12, 2004
11. Feb 12, 2004

### matt grime

forget that last post if you saw it, i absolutely apoligize, i misread your post.

Last edited: Feb 12, 2004
12. Feb 12, 2004

### NateTG

Don't sweat it, I've posted some beauties myself. I should be less adverserial in my response though. ;)

13. Feb 19, 2004

### MathNerd

Integer solutions of $$xy = x + y$$ where $$x,y \epsilon Z$$

$$xy - x - y = 0$$

$$(x-1)(y-1) = 1$$

let $$x' = x -1$$ and $$y' = y - 1$$ so $$x',y' \epsilon Z$$

So $$x' = 1 / y'$$

The only integers whose reciprocals are also integers are 1 and -1

So $$y' = 1$$ and $$x' = 1$$

So $$y = 2$$ and $$x = 2$$

AND

So $$y' = -1$$ and $$x' = -1$$

So $$y = 0$$ and $$x = 0$$

Therefore there are only two solutions to this diophantine equation $$x = y = 2$$ and $$x = y = 0$$

Last edited by a moderator: Feb 19, 2004