Prime factorization of very large integer with quadratic residue and its square roots
$begingroup$
We have a very large modulus integer $n$ and we have very large number $y$. We know that $y$ is a quadratic residue modulo $n$. Also we know all $4$ square roots of $y$.
What is the best way of prime factorization of $n$ ?
number-theory prime-factorization quadratic-residues
edited Dec 18 '18 at 13:31
Klangen
1,75811334
asked Oct 19 '18 at 18:19
l0veisreall0veisreal
11
$endgroup$
add a comment |
$begingroup$
We have a very large modulus integer $n$ and we have very large number $y$. We know that $y$ is a quadratic residue modulo $n$. Also we know all $4$ square roots of $y$.
What is the best way of prime factorization of $n$ ?
number-theory prime-factorization quadratic-residues
edited Dec 18 '18 at 13:31
Klangen
1,75811334
asked Oct 19 '18 at 18:19
l0veisreall0veisreal
11
$endgroup$
add a comment |
$begingroup$
We have a very large modulus integer $n$ and we have very large number $y$. We know that $y$ is a quadratic residue modulo $n$. Also we know all $4$ square roots of $y$.
What is the best way of prime factorization of $n$ ?
number-theory prime-factorization quadratic-residues
edited Dec 18 '18 at 13:31
Klangen
1,75811334
asked Oct 19 '18 at 18:19
l0veisreall0veisreal
11
$endgroup$
We have a very large modulus integer $n$ and we have very large number $y$. We know that $y$ is a quadratic residue modulo $n$. Also we know all $4$ square roots of $y$.
What is the best way of prime factorization of $n$ ?
number-theory prime-factorization quadratic-residues
number-theory prime-factorization quadratic-residues
edited Dec 18 '18 at 13:31
Klangen
1,75811334
asked Oct 19 '18 at 18:19
l0veisreall0veisreal
11
edited Dec 18 '18 at 13:31
Klangen
1,75811334
asked Oct 19 '18 at 18:19
l0veisreall0veisreal
11
edited Dec 18 '18 at 13:31
Klangen
1,75811334
edited Dec 18 '18 at 13:31
Klangen
1,75811334
edited Dec 18 '18 at 13:31
Klangen
1,75811334
1,75811334
asked Oct 19 '18 at 18:19
l0veisreall0veisreal
11
asked Oct 19 '18 at 18:19
l0veisreall0veisreal
11
asked Oct 19 '18 at 18:19
l0veisreall0veisreal
11
11
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If $a^2equiv b^2pmod N$, but $anotequiv bpmod N$, then $g=gcd(a-b,N)$
is a factor of $N$ with $1<g<N$.
answered Oct 19 '18 at 18:23
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
$begingroup$
so in my case y is b and a are square roots of y ???( i guess not)
$endgroup$
– l0veisreal
Oct 19 '18 at 18:49
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $a^2equiv b^2pmod N$, but $anotequiv bpmod N$, then $g=gcd(a-b,N)$
is a factor of $N$ with $1<g<N$.
answered Oct 19 '18 at 18:23
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
$begingroup$
so in my case y is b and a are square roots of y ???( i guess not)
$endgroup$
– l0veisreal
Oct 19 '18 at 18:49
add a comment |
$begingroup$
If $a^2equiv b^2pmod N$, but $anotequiv bpmod N$, then $g=gcd(a-b,N)$
is a factor of $N$ with $1<g<N$.
answered Oct 19 '18 at 18:23
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
$begingroup$
so in my case y is b and a are square roots of y ???( i guess not)
$endgroup$
– l0veisreal
Oct 19 '18 at 18:49
add a comment |
$begingroup$
If $a^2equiv b^2pmod N$, but $anotequiv bpmod N$, then $g=gcd(a-b,N)$
is a factor of $N$ with $1<g<N$.
answered Oct 19 '18 at 18:23
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
If $a^2equiv b^2pmod N$, but $anotequiv bpmod N$, then $g=gcd(a-b,N)$
is a factor of $N$ with $1<g<N$.
answered Oct 19 '18 at 18:23
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Oct 19 '18 at 18:23
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Oct 19 '18 at 18:23
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Oct 19 '18 at 18:23
Lord Shark the UnknownLord Shark the Unknown
105k1160132
105k1160132
$begingroup$
so in my case y is b and a are square roots of y ???( i guess not)
$endgroup$
– l0veisreal
Oct 19 '18 at 18:49
add a comment |
$begingroup$
so in my case y is b and a are square roots of y ???( i guess not)
$endgroup$
– l0veisreal
Oct 19 '18 at 18:49
$begingroup$
so in my case y is b and a are square roots of y ???( i guess not)
$endgroup$
– l0veisreal
Oct 19 '18 at 18:49
$begingroup$
so in my case y is b and a are square roots of y ???( i guess not)
$endgroup$
– l0veisreal
Oct 19 '18 at 18:49
add a comment |
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