Fun with p and qby Ilya Levin, Literatecode

Approved for public release; distribution unlimited

N = pq 2 < p ≠ q

0 < p < √ < q < N

0

N√p

q

Not much funin “somewhere”

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x2 mod N

N + 1 h = ––––––

2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1 4 9 16 4 15 7 1 18 16 16 18 1 7 15 4 16 9 4 1

√ h

N = 21

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1 4 9 16 4 15 7 1 18 16 16 18 1 7 15 4 16 9 4 1

√ b h d

N = 21

b2 = b d2 = d

|b-d|2 = 1(mod N)

b = xp d = yq

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THERE IS A SYMMETRY!

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Δh,b ≡ Δd,h

0123456789101112131415161718192021

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

0 N = 527 h = 264

b = 187

d = 341gcd(527,187) = 17 gcd(527, 341) = 31 527 = 17 × 31

N = 2046587

b = 984741 d = 1061847

p = gcd(d, N) = 929 q = gcd(b, N) = 2203

N = 133

b = 77 d = 57

p = gcd(b, N) = 7 q = gcd(d, N) = 19

Instead of p and q, why don't we start looking for b and d ?

Let the fun begin!

Braveheart