Given x

^{3}- Dy

^{3}= 1 I'll try to find modular solutions for x and y modulo D-1:

x

^{3}- Dy

^{3}= 1 mod D-1, and D mod D-1 = 1, so:

x

^{3}- y

^{3}= 1 mod D-1

Which is: (x-y)(x

^{2}+ xy + y

^{2}) = 1 mod D-1, and letting r be any residue with an inverse modulo D-1:

x-y = r mod D-1 and x

^{2}+ xy + y

^{2}= r

^{-1}mod D-1, so, x = y+r mod D-1, allowing me to substitute:

(y+r)

^{2}+ (y+r)y + y

^{2}= r

^{-1}mod D-1

and expanding out gives:

y

^{2}+ 2yr + r

^{2}+ y

^{2}+ ry + y

^{2}= r

^{-1}mod D-1, so:

3y

^{2}+ 3yr + r

^{2}- r

^{-1}= 0 mod D-1, and I'll use fractions--though it can be done without them--to make things easier for me and complete the square:

y

^{2}+ yr + r

^{2}/4 - r

^{2}/4+( r

^{2}- r

^{-1})/3 = 0 mod D-1, which is: (y+r/2)

^{2}= -(r

^{2}- 4r

^{-1})/12 mod D-1, so:

3(2y+r)

^{2}= 4r

^{-1 }- r

^{2}mod D-1

And I think it is kind of ugly. But I'd think you could get answers with it, but just have to be able to find a quadratic residue, if it exists.

Just a curiosity. I'll put this up and maybe come back later to check my derivation and try it with some numbers.

I don't know of a practical use for this result, but it seemed like a fun thing to just extend a bit, and see what I got.

James Harris

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