# Integer solution

1. Sep 7, 2010

### annoymage

1. The problem statement, all variables and given/known data

let $$x,y \in N$$. Solve the equation $$\frac{1}{x}+\frac{1}{y}=1$$

2. Relevant equations

n/a

3. The attempt at a solution

so i can see x=y=2 is the solution, hmm, isn't there any other solution.

and, this is one of three of my assignment question for 15% continuous assessment, does it seem too easy? should i prove that they are no other solution other than x=y=2?? help

Last edited: Sep 7, 2010
2. Sep 7, 2010

### TachyonRunner

You shouldn't prove that there is no other solution because there are other solutions. How about x = 3 and y = (3/2)? or x = 6 and y = (5/6) ??? This is an implicit function describing a curve (in this case a straight line) of some sort, which means there are infinite points (values of x and y) that satisfy the equation. Try making it an explicit function by solving for y (or x if you want) and then see what you get.

3. Sep 7, 2010

### annoymage

owh i forgot to wrote it, it's on the title "integer solution" hmm lemme edit

4. Sep 7, 2010

### Petek

Hint: If x > 2, then $\frac{1}{x}$ is ... .

Petek

5. Sep 7, 2010

### annoymage

If $$x>2$$ then $$\frac{1}{x}+\frac{1}{y}<\frac{1}{2}+\frac{1}{y}$$ and if $$y \geq 2$$, then $$\frac{1}{x}+\frac{1}{y}<\frac{1}{2}+\frac{1}{2}=1$$

similarly if $$y>2$$ and $$x \geq 2$$, so either case is not a solution. should i do like that?

6. Sep 7, 2010

### Petek

Yes, that's the right idea.

Petek

7. Sep 7, 2010

### snipez90

Petek illustrates an important method, but this diophantine equation can be solved explicitly. Simply clear denominators and rearrange to get xy - x - y = 0. Now try to add a constant to both sides so that you can factor the left hand side as a product of two linear factors, similar to completing the square.

8. Sep 8, 2010

### annoymage

cool, i get

$$(x-1)(y-1)=1$$ implies $$(x-1)=1 \ and\ (y-1)=1$$

but what if

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$$, i can't solve it explicitly can't i?

so i know$$x=y=z=3$$ is a solution

and i need to verify, $$x \in N-(3)\ ,y \in N-(3)\ ,z \in N-(3)$$ is not a solution

so for $$x=1,2$$ easy to verify

if $$x>3 \ \ \ \$$ $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<\frac{1}{3}+\frac{1}{y}+\frac{1}{z}$$ so when y=1,z=2 and y=2,z=1 and y=1,z>3 and y=2,z>3 and y>3,z=1 and y>3,z=2, is not a solution(i have to verify all these cases?)

and then repeat this tedious cases for y>3 and z>3

is that the way?

9. Sep 8, 2010

### Hurkyl

Staff Emeritus
Actually, 1 factors in two different ways over the integers:
1 = 1*1
1 = -1 * -1

10. Sep 8, 2010

### annoymage

owh yes, thank ;P

so (x-1)=-1 =>x=0 similarly y=0, which is not a solution.

is this argument correct? T_T

11. Sep 8, 2010

### hunt_mat

If
$$\frac{1}{x}+\frac{1}{y}=1$$
Then:
$$y=\frac{x}{x-1}$$
As y is an integer, we know that the above fraction is an interger, so x-1 must perfectly divide x, when can this happen?

12. Sep 8, 2010

### annoymage

hmm since gcd(x-1,x)=1 so x/(x-1) to be integer means x-1 must equal to 1 or -1, right?

hey, if i solve explicitly like this or like

that's the only solution right? and other integer is not a solution?? and i don't need to verify others is not a solution right?? right right??? is it right?

13. Sep 8, 2010

### hunt_mat

Correct, but you can get rid of one of the options with a bit of thought. I was mainly thinking that if x is an integer then the greatest divisor must be x/2 if x is even or less if x is odd.

14. Sep 8, 2010

### annoymage

wait what you mean by
?????

and thanks for this clue
i'm interesting to prove that statement ;P

Last edited: Sep 8, 2010
15. Sep 8, 2010

### hunt_mat

I always loathed number theory. Look at the what x-1=1 and x-1=-1 and see what it means with the original equation.

As with the rule, it was just an observation. I have no idea how to prove it.
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