- #1

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[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.

- Thread starter recon
- Start date

- #1

- 399

- 1

[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.

- #2

- 998

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[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?

- #3

- 399

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*smacks head*

Of course, consecutive numbers cannot be squares.

Of course, consecutive numbers cannot be squares.

- #4

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- #5

- 399

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Oh yeah, I phrased it wrongly.

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.

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:

- #6

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I'll check back in the morning~

- #7

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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.

- #8

- 998

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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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