Proving No Positive Integer Solutions for 1/x^2 + 1/xy + 1/y^2 = 1

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Homework Help Overview

The discussion revolves around proving that there are no positive integer solutions for the equation 1/x² + 1/xy + 1/y² = 1. Participants are exploring the implications of this equation and examining the properties of integers involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss rearranging the original equation and its implications, with one suggesting a contradiction arises from the derived form. Others reflect on properties of consecutive integers and their relationship to square numbers.

Discussion Status

The discussion is ongoing, with participants sharing insights and attempting to clarify concepts related to the problem. Some guidance has been offered regarding the properties of integers, but no consensus has been reached on a definitive proof.

Contextual Notes

Participants are constrained by the requirement to find positive integer solutions, and there is an acknowledgment that zero cannot be included in the solutions.

recon
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Prove that there are no positive integers x & y such that

[tex]\frac{1}{x^2} + \frac{1}{xy} + \frac{1}{y^2} = 1[/tex].

I really have no idea where to start. I tried rearranging the equation, but I could not see anything useful in any of the rearrangements.
 
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[tex]\frac{1}{x^2} + \frac{1}{xy} + \frac{1}{y^2} = 1[/tex]

[tex]\Longleftrightarrow x^2 + xy + y^2 = x^2y^2[/tex]

[tex]\Longleftrightarrow x^2 + 2xy + y^2 = (x+y)^2 = x^2y^2 + xy =xy(xy+1)[/tex]

ie. it implies

[tex](x+y)^2 = xy(xy+1)[/tex]

see a contradiction? :smile:
 
*smacks head*

Of course, consecutive numbers cannot be squares.
 
Right, products of consecutive positive integers can't be squares (I assume that's what you meant~). See an easy proof of that?
 
Oh yeah, I phrased it wrongly. :redface:

An easy proof? No, I can't find an EASY one, but this is the best I can do:

a-1, a and a+1 are consecutive integers.

[tex](a - 1)^2 = a^2 - 2a + 1[/tex]
[tex](a + 1)^2 = a^2 + 2a + 1[/tex]

a^2 - a and this is bigger than a^2 - 2a + 1 but smaller than a^2 + 2a + 1, since there is only square number between these two boundaries, a^2, this means that a * (a - 1) cannot be a square number.

The same reasoning, when applied to a * (a + 1), and using a + 2 reveals that it can't be a square number too.
 
Last edited:
I can't see most of that post right now, and I'm too tired to think at the moment :smile:

I'll check back in the morning~
 
[tex](x+y)^2 = xy(xy+1)[/tex]
can be a square number if x=y=0.
[tex]a=xy(xy+1)[/tex], i.e.[tex]a=z(z+1)[/tex]
So, if a is in this form, it can be a square number.
However, referring to the case, x cannot be 0 and y cannot be 0 neither.
 
Yep, if one of them is zero it's okay. As you mentioned, in this case both of them are positive (ie. nonzero), though.

Your proof is fine, recon, though you can make it simpler:

If [itex]a>0[/itex] then

[tex]a^2 \leq a^2 + a = a(a+1) \leq a^2 + 2a + 1 = (a+1)^2[/tex]

so if [itex]a(a+1) = b^2[/itex] then [itex]a<b<a+1[/itex], but there's no integer satisfying this if [itex]a[/itex] is an integer.
 

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