Figured out a way to calculate the modular inverse of some residue r modulo some integer N, by leveraging a modular factorization. Will give the working system, show an example then give the derivation.
System:
r-1 = (n-1)(r + 2my0) - 2md mod N
Where y0 is chosen as is m, with m not equal to r, and n and d are to be determined. They are found from:
2mdF0 = [F0(n-1) - 1](r + 2my0) mod N
and
F0 = r(r+2my0) mod N
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Example: Calculate the modular inverse of 11 modulo 137.
With N = 137, I will use m = 12, as: 7 = 144 mod 137
And looking at simple will use y0 = 1, so:
F0 = 11(11+2(12)(1)) = 11(35) = 111 mod 137
61d = 49n - 84 = 7(7n - 12) mod 137
If we now let d = 7d', get to 61d' = 7n - 12 mod 137, and n = -7 works with d' = -1.
So d = - 7, with n = -7, giving r-1 = (-7-1)(11+2(12)(1)) - 2(12)(-7) mod 137
So: r-1 = -6 + 31 = 25 mod 137, and 11(25) = 1 mod 137 as required.
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Derivation:
Given, some residue r modulo N whose modular inverse is to be determined, find D such that: m2 = D mod N, and m does not equal r.
Then for some x2 - Dy2 = F, I have the modular factorization modulo N:
(x-my)(x+my) = x2 - Dy2 = F mod N, and can set:
x - my = r mod N, then x+my =Fr-1 mod N.
Solving for x, with x = my + r mod N, and substituting into x2 - Dy2 = F gets me:
r(r+2my) = F mod N
So I can look at a y0 which will give an F0, and a difference of d, such that:
y = y0 + d
And subtract that F from a multiple n, of my initial F, substituting for y:
nF0 - F = r[(n-1)(r+2my0) - 2md] mod N
Now consider, nF0 - F = 1 mod N, which gives:
r[(n-1)(r+2my0) - 2md] = 1 mod N
And can solve from there for the modular inverse:
r-1 = (n-1)(r + 2my0) - 2md mod N
Now can substitute into x+my =Fr-1 mod N, using x = my + r mod N, and simplify somewhat to get the control equation:
2mdF0 = [F0(n-1) - 1](r + 2my0) mod N
and from before I have to show the full system as given:
F0 = r(r+2my0) mod N
Derivation complete.
James Harris
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Want a PDF? You can get one at my math group. There is a post there with a PDF attached.
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