How to show the relationship between the two sequences in Pollard's $rho$ factoring algorithm?
$begingroup$
$DeclareMathOperator{mdc}{gcd}$This is about Pollard's $rho$ factoring algorithm. I'd like to show the relationship between the sequence of numbers modulo $p$ and the sequence of numbers modulo $N$, the composite we wish to factor.
My shot at the problem is this possible theorem below. I'm not convinced it's all correct.
Theorem. Let $N=pq:$ be the product of two primes $p, q$. Let $(N_k)$ be a sequence of random numbers reduced modulo $N$. Let $(P_k)$ be a sequence defined by reducing each element of $(N_k)$ modulo $p$. If there exists natural numbers $i, j$ such that $i neq j$ and $p_i = p_j in (P_k)$ then it follows that
$$1 < mdc(|n_i - n_j|, N) leq N,$$ where $n_i, n_j in (N_k).$
Proof. Suppose there is $i,j$ naturals such that $i neq j$ and $p_i = p_j in (P_k)$. Divide both $n_i$ and $n_j$ by $p.$ By the division algorithm, we get
begin{align*}
n_i &= q_1 p + p_i\
n_j &= q_2 p + p_j.
end{align*}
Since $p_i = p_j,$ we may rewrite these equations as
begin{align*}
n_i &= q_1 p + p_i\
n_j &= q_2 p + p_i.
end{align*}
Subtracting the second equation from the first, we get
begin{align*}
n_i - n_j &= (q_1 p + p_i) -(q_2 p + p_i)\
&= q_1 p - q_2 p\
&= (q_1 - q_2) p,
end{align*}
implying $p$ is a factor of $n_i - n_j$ e hence of $|n_i - n_j|$.
Since $p$ is prime, then it's clear that $p > 1$. Therefore, $1 < mdc(|n_i-n_j|,N) leq N$, as desired. QED.
My main concern is whether the result is correct. I'll appreciate any style comments regarding clarity and elegance. Thank you!
proof-verification proof-writing alternative-proof
asked Jan 7 at 19:31
Luitpold AmbreLuitpold Ambre
135
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperator{mdc}{gcd}$This is about Pollard's $rho$ factoring algorithm. I'd like to show the relationship between the sequence of numbers modulo $p$ and the sequence of numbers modulo $N$, the composite we wish to factor.
My shot at the problem is this possible theorem below. I'm not convinced it's all correct.
Theorem. Let $N=pq:$ be the product of two primes $p, q$. Let $(N_k)$ be a sequence of random numbers reduced modulo $N$. Let $(P_k)$ be a sequence defined by reducing each element of $(N_k)$ modulo $p$. If there exists natural numbers $i, j$ such that $i neq j$ and $p_i = p_j in (P_k)$ then it follows that
$$1 < mdc(|n_i - n_j|, N) leq N,$$ where $n_i, n_j in (N_k).$
Proof. Suppose there is $i,j$ naturals such that $i neq j$ and $p_i = p_j in (P_k)$. Divide both $n_i$ and $n_j$ by $p.$ By the division algorithm, we get
begin{align*}
n_i &= q_1 p + p_i\
n_j &= q_2 p + p_j.
end{align*}
Since $p_i = p_j,$ we may rewrite these equations as
begin{align*}
n_i &= q_1 p + p_i\
n_j &= q_2 p + p_i.
end{align*}
Subtracting the second equation from the first, we get
begin{align*}
n_i - n_j &= (q_1 p + p_i) -(q_2 p + p_i)\
&= q_1 p - q_2 p\
&= (q_1 - q_2) p,
end{align*}
implying $p$ is a factor of $n_i - n_j$ e hence of $|n_i - n_j|$.
Since $p$ is prime, then it's clear that $p > 1$. Therefore, $1 < mdc(|n_i-n_j|,N) leq N$, as desired. QED.
My main concern is whether the result is correct. I'll appreciate any style comments regarding clarity and elegance. Thank you!
proof-verification proof-writing alternative-proof
asked Jan 7 at 19:31
Luitpold AmbreLuitpold Ambre
135
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperator{mdc}{gcd}$This is about Pollard's $rho$ factoring algorithm. I'd like to show the relationship between the sequence of numbers modulo $p$ and the sequence of numbers modulo $N$, the composite we wish to factor.
My shot at the problem is this possible theorem below. I'm not convinced it's all correct.
Theorem. Let $N=pq:$ be the product of two primes $p, q$. Let $(N_k)$ be a sequence of random numbers reduced modulo $N$. Let $(P_k)$ be a sequence defined by reducing each element of $(N_k)$ modulo $p$. If there exists natural numbers $i, j$ such that $i neq j$ and $p_i = p_j in (P_k)$ then it follows that
$$1 < mdc(|n_i - n_j|, N) leq N,$$ where $n_i, n_j in (N_k).$
Proof. Suppose there is $i,j$ naturals such that $i neq j$ and $p_i = p_j in (P_k)$. Divide both $n_i$ and $n_j$ by $p.$ By the division algorithm, we get
begin{align*}
n_i &= q_1 p + p_i\
n_j &= q_2 p + p_j.
end{align*}
Since $p_i = p_j,$ we may rewrite these equations as
begin{align*}
n_i &= q_1 p + p_i\
n_j &= q_2 p + p_i.
end{align*}
Subtracting the second equation from the first, we get
begin{align*}
n_i - n_j &= (q_1 p + p_i) -(q_2 p + p_i)\
&= q_1 p - q_2 p\
&= (q_1 - q_2) p,
end{align*}
implying $p$ is a factor of $n_i - n_j$ e hence of $|n_i - n_j|$.
Since $p$ is prime, then it's clear that $p > 1$. Therefore, $1 < mdc(|n_i-n_j|,N) leq N$, as desired. QED.
My main concern is whether the result is correct. I'll appreciate any style comments regarding clarity and elegance. Thank you!
proof-verification proof-writing alternative-proof
asked Jan 7 at 19:31
Luitpold AmbreLuitpold Ambre
135
$endgroup$
$DeclareMathOperator{mdc}{gcd}$This is about Pollard's $rho$ factoring algorithm. I'd like to show the relationship between the sequence of numbers modulo $p$ and the sequence of numbers modulo $N$, the composite we wish to factor.
My shot at the problem is this possible theorem below. I'm not convinced it's all correct.
Theorem. Let $N=pq:$ be the product of two primes $p, q$. Let $(N_k)$ be a sequence of random numbers reduced modulo $N$. Let $(P_k)$ be a sequence defined by reducing each element of $(N_k)$ modulo $p$. If there exists natural numbers $i, j$ such that $i neq j$ and $p_i = p_j in (P_k)$ then it follows that
$$1 < mdc(|n_i - n_j|, N) leq N,$$ where $n_i, n_j in (N_k).$
Proof. Suppose there is $i,j$ naturals such that $i neq j$ and $p_i = p_j in (P_k)$. Divide both $n_i$ and $n_j$ by $p.$ By the division algorithm, we get
begin{align*}
n_i &= q_1 p + p_i\
n_j &= q_2 p + p_j.
end{align*}
Since $p_i = p_j,$ we may rewrite these equations as
begin{align*}
n_i &= q_1 p + p_i\
n_j &= q_2 p + p_i.
end{align*}
Subtracting the second equation from the first, we get
begin{align*}
n_i - n_j &= (q_1 p + p_i) -(q_2 p + p_i)\
&= q_1 p - q_2 p\
&= (q_1 - q_2) p,
end{align*}
implying $p$ is a factor of $n_i - n_j$ e hence of $|n_i - n_j|$.
Since $p$ is prime, then it's clear that $p > 1$. Therefore, $1 < mdc(|n_i-n_j|,N) leq N$, as desired. QED.
My main concern is whether the result is correct. I'll appreciate any style comments regarding clarity and elegance. Thank you!
proof-verification proof-writing alternative-proof
proof-verification proof-writing alternative-proof
asked Jan 7 at 19:31
Luitpold AmbreLuitpold Ambre
135
asked Jan 7 at 19:31
Luitpold AmbreLuitpold Ambre
135
edited Jan 7 at 19:48
Luitpold Ambre
asked Jan 7 at 19:31
Luitpold AmbreLuitpold Ambre
135
asked Jan 7 at 19:31
Luitpold AmbreLuitpold Ambre
135
asked Jan 7 at 19:31
Luitpold AmbreLuitpold Ambre
135
135
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