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Wednesday, May 23, 2012

Cubic Diophantine computations

So yesterday I found myself wondering what would happen if I extended my idea of looking modulo D-1 to a cubic Diophantine, which is just about curiosity! Here, I'll do a post with some computations.

To recap, I found that given x3 - Dy3 = 1, I could solve for y with:

3(2y+r)2  = 4r-1 - r2  mod D-1

where x = y+r mod D-1, and r is a residue which has a modular inverse modulo D-1.

So I'll arbitrarily use a situation where I have a solution, which is D=26, as x=3, y=1 works.

Then:  3(2y+r)2  = 4r-1 - r2  mod 25, and I'll try r=1, so (2y+1)2  = 1 mod 25.

Which gives 2y + 1 = 1 or -1 mod 25. So y = 0 or -1 mod 25, which works trivially.

Moving to r=2.

3(2y+2)2  = 4(13) - (2)2  mod 25 = 23 mod 25, so (2y+2)2  = 16 mod 25.

Which gives two solutions. One is 2y + 2 = 4 mod 25, so y = 1 mod 25.

For the other 2y = -4 mod 25, so y = 22 mod 25, then x = 24 mod 25.

And (24)3 - 26(22)3 = -263024 = 1 - (10521)(25) = 1 mod 25 as required.


James Harris
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