Mixing
Sum of 2 squares implies efficient factorization
$begingroup$
I'm concerning myself with factoring semi-primes and believe that if given a large semi-prime ($N$) one finds a non-trivial sum of squares representation:
$$ x^2 + y^2 = N$$
Then one can efficiently retrieve a factorization of $N$. My intuition for this belief is that I think the "factoring" problem over the gaussian integers is difficult and note that such a representation can be in linear time converted into a factorization $(x + yi) ( x- yi)$ when considering gaussian integers.
But it's not clear to me how to leverage this information for regular integers.
More Notes:
So I realized after @Jack D'Aruzio's comment that this information could be exploited by computing $i$ mod $N$
That is finding some constant $c$ such that $c^2 = -1 mod N$. Then we have that
$$ (x + cy)(x - cy) equiv 0 mod N$$
And therefore at least one of $(x + cy)$ and $(x - cy)$ must be a zero divisor and can be used to recover factors of (assuming we don't have the trivial, non-trivial pair) $N$.
Of course there is no guarantee that we can find such a $c$, and even if its guaranteed to exist such a $c$ might be computationally as expensive to find as factoring itself
number-theory elementary-number-theory algebraic-number-theory factoring cryptography
asked Jan 2 at 21:47
frogeyedpeasfrogeyedpeas
7,44071949
$endgroup$
add a comment |
$begingroup$
I'm concerning myself with factoring semi-primes and believe that if given a large semi-prime ($N$) one finds a non-trivial sum of squares representation:
$$ x^2 + y^2 = N$$
Then one can efficiently retrieve a factorization of $N$. My intuition for this belief is that I think the "factoring" problem over the gaussian integers is difficult and note that such a representation can be in linear time converted into a factorization $(x + yi) ( x- yi)$ when considering gaussian integers.
But it's not clear to me how to leverage this information for regular integers.
More Notes:
So I realized after @Jack D'Aruzio's comment that this information could be exploited by computing $i$ mod $N$
That is finding some constant $c$ such that $c^2 = -1 mod N$. Then we have that
$$ (x + cy)(x - cy) equiv 0 mod N$$
And therefore at least one of $(x + cy)$ and $(x - cy)$ must be a zero divisor and can be used to recover factors of (assuming we don't have the trivial, non-trivial pair) $N$.
Of course there is no guarantee that we can find such a $c$, and even if its guaranteed to exist such a $c$ might be computationally as expensive to find as factoring itself
number-theory elementary-number-theory algebraic-number-theory factoring cryptography
asked Jan 2 at 21:47
frogeyedpeasfrogeyedpeas
7,44071949
$endgroup$
$begingroup$
You are just re-discovering the key ingredients of the quadratic sieve and of the general field sieve.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:12
$begingroup$
The point is that the factorization problem in $mathbb{Z}$ and $mathbb{Z}[i]$ is just as difficult, but if $N$ is represented by some binary quadratic form (with a negative discriminant) (and a small class number), $x$ and $y$ have to fulfill many arithmetic constraints, so they can be sifted by exploiting the informations provided by Legendre/Jacobi/Kronecker symbols.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:18
2
$begingroup$
meanwhile, if you get two different expressions as sum of two squares, you get a factoring. zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
$endgroup$
– Will Jagy
Jan 2 at 22:58
$begingroup$
you may want to look at this method using the sum of 2 squares to factor numbers. It is different from Euler's method. math.stackexchange.com/questions/3066531/…
$endgroup$
– user25406
Jan 9 at 0:22
add a comment |
$begingroup$
I'm concerning myself with factoring semi-primes and believe that if given a large semi-prime ($N$) one finds a non-trivial sum of squares representation:
$$ x^2 + y^2 = N$$
Then one can efficiently retrieve a factorization of $N$. My intuition for this belief is that I think the "factoring" problem over the gaussian integers is difficult and note that such a representation can be in linear time converted into a factorization $(x + yi) ( x- yi)$ when considering gaussian integers.
But it's not clear to me how to leverage this information for regular integers.
More Notes:
So I realized after @Jack D'Aruzio's comment that this information could be exploited by computing $i$ mod $N$
That is finding some constant $c$ such that $c^2 = -1 mod N$. Then we have that
$$ (x + cy)(x - cy) equiv 0 mod N$$
And therefore at least one of $(x + cy)$ and $(x - cy)$ must be a zero divisor and can be used to recover factors of (assuming we don't have the trivial, non-trivial pair) $N$.
Of course there is no guarantee that we can find such a $c$, and even if its guaranteed to exist such a $c$ might be computationally as expensive to find as factoring itself
number-theory elementary-number-theory algebraic-number-theory factoring cryptography
asked Jan 2 at 21:47
frogeyedpeasfrogeyedpeas
7,44071949
$endgroup$
I'm concerning myself with factoring semi-primes and believe that if given a large semi-prime ($N$) one finds a non-trivial sum of squares representation:
$$ x^2 + y^2 = N$$
Then one can efficiently retrieve a factorization of $N$. My intuition for this belief is that I think the "factoring" problem over the gaussian integers is difficult and note that such a representation can be in linear time converted into a factorization $(x + yi) ( x- yi)$ when considering gaussian integers.
But it's not clear to me how to leverage this information for regular integers.
More Notes:
So I realized after @Jack D'Aruzio's comment that this information could be exploited by computing $i$ mod $N$
That is finding some constant $c$ such that $c^2 = -1 mod N$. Then we have that
$$ (x + cy)(x - cy) equiv 0 mod N$$
And therefore at least one of $(x + cy)$ and $(x - cy)$ must be a zero divisor and can be used to recover factors of (assuming we don't have the trivial, non-trivial pair) $N$.
Of course there is no guarantee that we can find such a $c$, and even if its guaranteed to exist such a $c$ might be computationally as expensive to find as factoring itself
number-theory elementary-number-theory algebraic-number-theory factoring cryptography
number-theory elementary-number-theory algebraic-number-theory factoring cryptography
asked Jan 2 at 21:47
frogeyedpeasfrogeyedpeas
7,44071949
asked Jan 2 at 21:47
frogeyedpeasfrogeyedpeas
7,44071949
edited Jan 3 at 16:24
frogeyedpeas
asked Jan 2 at 21:47
frogeyedpeasfrogeyedpeas
7,44071949
asked Jan 2 at 21:47
frogeyedpeasfrogeyedpeas
7,44071949
asked Jan 2 at 21:47
frogeyedpeasfrogeyedpeas
7,44071949
7,44071949
$begingroup$
You are just re-discovering the key ingredients of the quadratic sieve and of the general field sieve.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:12
$begingroup$
The point is that the factorization problem in $mathbb{Z}$ and $mathbb{Z}[i]$ is just as difficult, but if $N$ is represented by some binary quadratic form (with a negative discriminant) (and a small class number), $x$ and $y$ have to fulfill many arithmetic constraints, so they can be sifted by exploiting the informations provided by Legendre/Jacobi/Kronecker symbols.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:18
2
$begingroup$
meanwhile, if you get two different expressions as sum of two squares, you get a factoring. zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
$endgroup$
– Will Jagy
Jan 2 at 22:58
$begingroup$
you may want to look at this method using the sum of 2 squares to factor numbers. It is different from Euler's method. math.stackexchange.com/questions/3066531/…
$endgroup$
– user25406
Jan 9 at 0:22
add a comment |
$begingroup$
You are just re-discovering the key ingredients of the quadratic sieve and of the general field sieve.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:12
$begingroup$
The point is that the factorization problem in $mathbb{Z}$ and $mathbb{Z}[i]$ is just as difficult, but if $N$ is represented by some binary quadratic form (with a negative discriminant) (and a small class number), $x$ and $y$ have to fulfill many arithmetic constraints, so they can be sifted by exploiting the informations provided by Legendre/Jacobi/Kronecker symbols.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:18
2
$begingroup$
meanwhile, if you get two different expressions as sum of two squares, you get a factoring. zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
$endgroup$
– Will Jagy
Jan 2 at 22:58
$begingroup$
you may want to look at this method using the sum of 2 squares to factor numbers. It is different from Euler's method. math.stackexchange.com/questions/3066531/…
$endgroup$
– user25406
Jan 9 at 0:22
$begingroup$
You are just re-discovering the key ingredients of the quadratic sieve and of the general field sieve.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:12
$begingroup$
You are just re-discovering the key ingredients of the quadratic sieve and of the general field sieve.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:12
$begingroup$
The point is that the factorization problem in $mathbb{Z}$ and $mathbb{Z}[i]$ is just as difficult, but if $N$ is represented by some binary quadratic form (with a negative discriminant) (and a small class number), $x$ and $y$ have to fulfill many arithmetic constraints, so they can be sifted by exploiting the informations provided by Legendre/Jacobi/Kronecker symbols.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:18
$begingroup$
The point is that the factorization problem in $mathbb{Z}$ and $mathbb{Z}[i]$ is just as difficult, but if $N$ is represented by some binary quadratic form (with a negative discriminant) (and a small class number), $x$ and $y$ have to fulfill many arithmetic constraints, so they can be sifted by exploiting the informations provided by Legendre/Jacobi/Kronecker symbols.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:18
2
2
$begingroup$
meanwhile, if you get two different expressions as sum of two squares, you get a factoring. zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
$endgroup$
– Will Jagy
Jan 2 at 22:58
$begingroup$
meanwhile, if you get two different expressions as sum of two squares, you get a factoring. zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
$endgroup$
– Will Jagy
Jan 2 at 22:58
$begingroup$
you may want to look at this method using the sum of 2 squares to factor numbers. It is different from Euler's method. math.stackexchange.com/questions/3066531/…
$endgroup$
– user25406
Jan 9 at 0:22
$begingroup$
you may want to look at this method using the sum of 2 squares to factor numbers. It is different from Euler's method. math.stackexchange.com/questions/3066531/…
$endgroup$
– user25406
Jan 9 at 0:22
add a comment |
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$begingroup$
You are just re-discovering the key ingredients of the quadratic sieve and of the general field sieve.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:12
$begingroup$
The point is that the factorization problem in $mathbb{Z}$ and $mathbb{Z}[i]$ is just as difficult, but if $N$ is represented by some binary quadratic form (with a negative discriminant) (and a small class number), $x$ and $y$ have to fulfill many arithmetic constraints, so they can be sifted by exploiting the informations provided by Legendre/Jacobi/Kronecker symbols.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:18
2
$begingroup$
meanwhile, if you get two different expressions as sum of two squares, you get a factoring. zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf
$endgroup$
– Will Jagy
Jan 2 at 22:58
$begingroup$
you may want to look at this method using the sum of 2 squares to factor numbers. It is different from Euler's method. math.stackexchange.com/questions/3066531/…
$endgroup$
– user25406
Jan 9 at 0:22