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This is in relation to a problem dealing with the three-dimensional analogue of Pell's Equation. I would like to factor $ x^3+Dy^3+D^2z^3-3Dxyz $ into $\frac{1}{2}(x+Dy+D^2y)$ and another factor.

I would like just a hint on how to go about doing this as I'm pretty much stumped.

EDIT: I think the factor I was looking for is $\frac{1}{2}(x+Ny+N^2z)$ where $N=D^3$.

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Try long division? –  a little don Nov 9 '10 at 4:47
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2 Answers

up vote 7 down vote accepted

You've got the factor wrong. There is a well-known identity

$$a^3 + b^3 + c^3 - 3abc = \frac{1}{2} \left( a + b + c \right) \left( (a - b)^2 + (b - c)^2 + (c - a)^2 \right)$$

which you were probably thinking of, but to get the correct factorization you should take $a = x, b = \sqrt[3]{D} y, c = \sqrt[3]{D^2} z$.

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Thanks. I'll work with that and see what I get. –  James Nov 9 '10 at 23:09
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Thanks to Qiaochu Yuan's identity I was able to factor into

$ \frac{1}{2}(x+Ny+N^2z)[(x-Ny)^2+(Ny-N^2z)^2+(N^2z-x)^2] $

which shows that if

$ x^3+Dy^3+D^2z^3-3Dxyz=1 $ and $ x,y,z $ are very large positive, then

$ x-Dy $ and $ y-Dz $ must be small.

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