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• Factoring Integers via Diophantine Approximation

Shubham Sahai Srivastava

Indian Institute of Technology, Kanpur

ssahai@cse.iitk.ac.in

January 16, 2014

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 1 / 14

• Introduction and Surview

The task of factoring large composite integer N has a long history and is still a challenging problem.

Here, this task is reduced to the following diophantine approximation :

Definition (Problem)

Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

1. | ∑t

i=1 ei log pi − log N| ≤ N−cp o(1) t

2. | ∑t

i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

where, c > 1 and p1, ...pt are first t prime numbers.

Whats next ??

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 2 / 14

• Introduction and Surview

The task of factoring large composite integer N has a long history and is still a challenging problem.

Here, this task is reduced to the following diophantine approximation :

Definition (Problem)

Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

1. | ∑t

i=1 ei log pi − log N| ≤ N−cp o(1) t

2. | ∑t

i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

where, c > 1 and p1, ...pt are first t prime numbers.

Whats next ??

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 2 / 14

• Introduction and Surview

The task of factoring large composite integer N has a long history and is still a challenging problem.

Here, this task is reduced to the following diophantine approximation :

Definition (Problem)

Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

1. | ∑t

i=1 ei log pi − log N| ≤ N−cp o(1) t

2. | ∑t

i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

where, c > 1 and p1, ...pt are first t prime numbers.

Whats next ??

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 2 / 14

• Introduction and Surview

Given these t + 2 diophantine approximations of log N, we can factorize N as follows:

The integer u := ∏

ej>0 p ej j must be close approximation to vN, where

v := ∏

ej 1, β, γ ≥ 0 be fixed and let pt < N. If (e1, ..., et) ∈ Zt satisfies the inequalities

1. | ∑t

i=1 ei log pi − log N| ≤ N−cp β+o(1) t

2. | ∑t

i=1 ei log pi | ≤ (2c − 1) log N + 2δ log pt

then we have for u := ∏

ej>0 p ej j , v :=

∏ ej

• Introduction and Surview

Given these t + 2 diophantine approximations of log N, we can factorize N as follows:

The integer u := ∏

ej>0 p ej j must be close approximation to vN, where

v := ∏

ej 1, β, γ ≥ 0 be fixed and let pt < N. If (e1, ..., et) ∈ Zt satisfies the inequalities

1. | ∑t

i=1 ei log pi − log N| ≤ N−cp β+o(1) t

2. | ∑t

i=1 ei log pi | ≤ (2c − 1) log N + 2δ log pt

then we have for u := ∏

ej>0 p ej j , v :=

∏ ej

• Introduction and Surview

Given these t + 2 diophantine approximations of log N, we can factorize N as follows:

The integer u := ∏

ej>0 p ej j must be close approximation to vN, where

v := ∏

ej 1, β, γ ≥ 0 be fixed and let pt < N. If (e1, ..., et) ∈ Zt satisfies the inequalities

1. | ∑t

i=1 ei log pi − log N| ≤ N−cp β+o(1) t

2. | ∑t

i=1 ei log pi | ≤ (2c − 1) log N + 2δ log pt

then we have for u := ∏

ej>0 p ej j , v :=

∏ ej

• Introduction and Surview

So, we have |u − vN| ≤ p1+o(1)t

Hence, the residue u (mod N) factorizes completely over the primes p1, ..., pt

And we obtain a non-trivial congruence∏ ej>0

p ej j = ±

∏t j=1 p

ej j (mod N).

Given t + 2 of these congruences we compute x , y satisfying x2 = y2

(mod N)

So, we can compute a factor of N as gcd(x+y, N).

This gives us one factor and thus we can reduce N, by divinding N with this factor and continuing till we completely factorize N.

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 4 / 14

• Introduction and Surview

So, we have |u − vN| ≤ p1+o(1)t Hence, the residue u (mod N) factorizes completely over the primes p1, ..., pt

And we obtain a non-trivial congruence∏ ej>0

p ej j = ±

∏t j=1 p

ej j (mod N).

Given t + 2 of these congruences we compute x , y satisfying x2 = y2

(mod N)

So, we can compute a factor of N as gcd(x+y, N).

This gives us one factor and thus we can reduce N, by divinding N with this factor and continuing till we completely factorize N.

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 4 / 14

• Introduction and Surview

So, we have |u − vN| ≤ p1+o(1)t Hence, the residue u (mod N) factorizes completely over the primes p1, ..., pt

And we obtain a non-trivial congruence∏ ej>0

p ej j = ±

∏t j=1 p

ej j (mod N).

Given t + 2 of these congruences we compute x , y satisfying x2 = y2

(mod N)

So, we can compute a factor of N as gcd(x+y, N).

This gives us one factor and thus we can reduce N, by divinding N with this factor and continuing till we completely factorize N.

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 4 / 14

• Introduction and Surview

So, we have |u − vN| ≤ p1+o(1)t Hence, the residue u (mod N) factorizes completely over the primes p1, ..., pt

And we obtain a non-trivial congruence∏ ej>0

p ej j = ±

∏t j=1 p

ej j (mod N).

Given t + 2 of these congruences we compute x , y satisfying x2 = y2

(mod N)

So, we can compute a factor of N as gcd(x+y, N).

This gives us one factor and thus we can reduce N, by divinding N with this factor and continuing till we completely factorize N.

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 4 / 14

• Introduction and Surview

So, we have |u − vN| ≤ p1+o(1)t Hence, the residue u (mod N) factorizes completely over the primes p1, ..., pt

And we obtain a non-trivial congruence∏ ej>0

p ej j = ±

∏t j=1 p

ej j (mod N).

Given t + 2 of these congruences we compute x , y satisfying x2 = y2

(mod N)

So, we can compute a factor of N as gcd(x+y, N).

This gives us one factor and thus we can reduce N, by divinding N with this factor and continuing till we completely factorize N.

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 4 / 14

• Reduction to Lattice problem

So, we are good to go, if we are able to solve the following problem:

Definition (Diophantine Approximation Problem)

Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

1. | ∑t

i=1 ei log pi − log N| ≤ N−cp o(1) t

2. | ∑t

i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

where, c > 1 and p1, ...pt are first t prime numbers.

The above problem can be formulated as a nearly closest vector problem in the 1-norm.

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 5 / 14

• Reduction to Lattice problem

So, we are good to go, if we are able to solve the following problem:

Definition (Diophantine Approximation Problem)

Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

1. | ∑t

i=1 ei log pi − log N| ≤ N−cp o(1) t

2. | ∑t

i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

where, c > 1 and p1, ...pt are first t prime numbers.

The above problem can be formulated as a nearly closest vector problem in the 1-norm.

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 5 / 14

• Reduction to Lattice problem

So, we are good to go, if we are able to solve the following problem:

Definition (Diophantine Approximation Problem)

Find atleast t + 2 integer vectors (e1, e2, ...et) ∈ Zt satisfying:

1. | ∑t

i=1 ei log pi − log N| ≤ N−cp o(1) t

2. | ∑t

i=1 ei log pi | ≤ (2c − 1) log N + 2 log pt

where, c > 1 and p1, ...pt are first t prime numbers.

The above problem can be formulated as a nearly closest vector problem in the 1-norm.

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 5 / 14

• Reduction to Lattice problem

We associate with N a point N ∈ Rt+1

and with the primes p1, ..., pt a lattice L ⊂ Rt+1 of rank t and basis B.

B =

 log p1 0 · · · 0

... . . .

...

0 . . . 0

0 0 · · · log pt Nc log p1 N

c log p2 · · · Nc log pt

 , N = 

0 0 ... 0

Nc ln N ′

 , c ≥ 1

Shubham Sahai Srivastava (IITK) Factoring Integers January 16, 2014 6 / 14

• Reduction to Lattice problem

We associate with N a point N ∈ Rt+1 and with the primes p

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