Solving Integer Equation: 1/x + 1/y = 1

[annoymage]

Homework Statement

let [tex]x,y \in N[/tex]. Solve the equation [tex]\frac{1}{x}+\frac{1}{y}=1[/tex]

Homework Equations

n/a

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

 

[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.

 

[annoymage]

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

 

[Petek]

Hint: If x > 2, then [itex]\frac{1}{x}[/itex] is ... .

Petek

 

[annoymage]

If [tex]x>2[/tex] then [tex]\frac{1}{x}+\frac{1}{y}<\frac{1}{2}+\frac{1}{y}[/tex] and if [tex]y \geq 2[/tex], then [tex]\frac{1}{x}+\frac{1}{y}<\frac{1}{2}+\frac{1}{2}=1[/tex]

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

 

[Petek]

Yes, that's the right idea.

Petek

 

[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.

 

[annoymage]

cool, i get

[tex](x-1)(y-1)=1[/tex] implies [tex](x-1)=1 \ and\ (y-1)=1[/tex]

but what if

[tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1[/tex], i can't solve it explicitly can't i?

so i know[tex]x=y=z=3[/tex] is a solution

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

so for [tex]x=1,2[/tex] easy to verify

if [tex]x>3 \ \ \ \ [/tex] [tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<\frac{1}{3}+\frac{1}{y}+\frac{1}{z}[/tex] 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?

 

[Hurkyl]

cool, i get

[tex](x-1)(y-1)=1[/tex] implies [tex](x-1)=1 \ and\ (y-1)=1[/tex]

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

 

[annoymage]

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

owh yes, thank ;P

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

but how about this

cool, i get

[tex](x-1)(y-1)=1[/tex] implies [tex](x-1)=1 \ and\ (y-1)=1[/tex]

but what if

[tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1[/tex], i can't solve it explicitly can't i?

so i know[tex]x=y=z=3[/tex] is a solution

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

so for [tex]x=1,2[/tex] easy to verify

if [tex]x>3 \ \ \ \ [/tex] [tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<\frac{1}{3}+\frac{1}{y}+\frac{1}{z}[/tex] 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?

is this argument correct? T_T

 

[hunt_mat]

If
[tex]
\frac{1}{x}+\frac{1}{y}=1
[/tex]
Then:
[tex]
y=\frac{x}{x-1}
[/tex]
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?

 

[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

cool, i get

[tex](x-1)(y-1)=1[/tex] implies [tex](x-1)=1 \ and\ (y-1)=1[/tex]

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?

 

[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.

 

[annoymage]

wait what you mean by

you can get rid of one of the options with a bit of thought

?and thanks for this clue

if x is an integer then the greatest divisor must be x/2 if x is even or less if x is odd.

i'm interesting to prove that statement ;P

 

[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.