Sunday, February 03, 2008


JSH: Little congruence result

With non-zero coprime integers n_1 and n_2, if f_1 = r_1 mod n_1 and f_1 = r_2 mod n_2, you can find f_1 mod p_1*p_2 with

f_1 = r_1 + j*n_1 mod n_1*n_2

where j = (r_2 - r_1)*n_1^{-1} mod n_2.

I like this result better than the Chinese Remainder theorem, as you can just go by two's—and it's my discovery.

I like primarily using my own math.
Awesome little result where I find it mazing that no one has found it before as to me it's so much simpler to understand and use than the Chinese Remainder theorem, like notice, it's quite obvious and direct why n_1 MUST be coprime to n_2, as you need its modular inverse modulo n_2. And you can use my little congruence result with as many composites as you want and figure out immediately how many iterations it will take, as if n is the number of composites then ln n/ln 2 is the approximate number.

So to do 128 composites would require 7 iterations, taking your results by two's.

That little tidbit is just an extra though for this latest research path.

I like those little extras. Last little extra for me was my prime counting function coming in after I discovered my proof of Fermat's Last Theorem.

These tidbits are interesting in that they cover areas where mathematicians already have something, where I just find something better.

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