## Saturday, June 21, 2014

Pivotal to me and my math reality was one important teacher. Her name was Mrs. Boyd an algebra teacher in the Tift County school system she was the dreaded reality of arrival into the 7th grade. Her class was considered legendarily hard.

And it was hard for me, at first. She liked to give lots of homework including word problems. And I remember laboring over the weekend with a particularly long set. I think it was over a hundred. And something weird happened at some point, and suddenly every single one of them seemed easy.

I saw a pattern. Zipped through the rest of them, and suddenly algebra was ok. And I think there is something to having kids just do algebra, with as much instruction as you can of course, until they reach that magic moment. As I think it's just a matter of doing enough until the patterns emerge.

And then I felt myself to be in a different position in class and considered now with detached fascination as my classmates continued to struggle. One of my favorite memories was when I noticed a neat trick that would make a particular bit of algebra even easier. Mrs. Boyd listened to me, and if I remember correctly, put her finger to her lips, and called me to her desk.

She told me that was good but it might confuse the others.

It was like I was part of a special club. I could see it in her eyes. We were in the know.

James Harris

## Tuesday, June 17, 2014

You can focus on my cubic residue work in the opposite direction to see quadratic residue structure:

(2my+r)2  = 3-1(4Fr-1 - r2 ) mod N

if 3 is coprime to N, r is any residue modulo N for which its modular inverse exists,

where m3 = D mod N, and:

x3 - Dy3 = F, and r = x - my mod N.

If 3 is not coprime to N then you'd simply have:

(2my+r)2  = (4Fr-1 - r2 )/3 mod N

For example, let m=1 and D=1. And let x=2, and y = 1. Then F = 7. And we're set for any N. To use the first let's let N not have 3 as a factor then:

(r+2)2  = 3-1(28r-1 - r2 ) mod N, and r = 1 mod N

Which works.

The result comes from solving the cubic modulo N, and I realized you could also look in a different direction.

James Harris

## Friday, June 06, 2014

### Some cubic modular work

As much as I talk about binary quadratic forms turns out I also have extended concepts pioneered with them to cubic residues.

So I have found the following modular solution.

Given:  x3 - Dy3 = F

I can solve modulo N, where N is a cubic residue of D, m is the cubed residue, and r is a residue modulo N, where its modular inverse exists:

x = my + r mod N

and

3(2my+r)2  = 4Fr-1 - r2  mod N

If you wish to see it used, you can just check:

somemath.blogspot.com/2012/05/cubic-diophantine-computations.html

The ideas I use can be generalized further. But I think I went just to a cubic just to see what it might look like. Result seems very pure to me as in I'm not sure how useful it is.

James Harris