x + y + j + 1 = n

^{2}or 2n

^{2}

if x = 1 mod D, x

^{2}- Dy

^{2}= 1, and j = (x+Dy-1)/D

For example, 8

^{2}- 7*3

^{2}= 1, so x = 8, and 8 mod 7 = 1.

j = (8 + 7*3 - 1)/7 = (7 + 7*3)/7 = 4

8 + 3 + 4 + 1 = 16 = 4

^{2}

That's a simple example to make this post easy to write for me, but the result is true over infinity.

So whenever x = 1 mod D, with x

^{2}- Dy

^{2}= 1, then these rules are forced, absolutely.

Seem easy or trivial?

It's another key piece of the explanation for an ancient math mystery.

James Harris

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