When can we factor $N$ efficiently with a representation $N^2=a^2+b^2$?
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Here :
Can the sum of two squares be used to factor large numbers?
an idea to factor a large number $N=a^2+b^2$ is shown.
Under which conditions does a representation $$N^2=u^2+v^2$$ exist , such that $gcd(N,gcd(u,v))$ is a non-trivial factor of $N$ ?
I guess this is the case, when $N$ has at least one prime factor of the form $4k+1$ and is not a prime power, but I am not sure whether this is true and how it can be proven.
number-theory elementary-number-theory prime-numbers square-numbers
asked Jan 9 at 15:49
PeterPeter
48.1k1139133
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add a comment |
$begingroup$
Here :
Can the sum of two squares be used to factor large numbers?
an idea to factor a large number $N=a^2+b^2$ is shown.
Under which conditions does a representation $$N^2=u^2+v^2$$ exist , such that $gcd(N,gcd(u,v))$ is a non-trivial factor of $N$ ?
I guess this is the case, when $N$ has at least one prime factor of the form $4k+1$ and is not a prime power, but I am not sure whether this is true and how it can be proven.
number-theory elementary-number-theory prime-numbers square-numbers
asked Jan 9 at 15:49
PeterPeter
48.1k1139133
$endgroup$
1
$begingroup$
To enumerate all the ways such that $M = u^2+v^2$ It suffices to know that for $k$ the order of $p bmod 4$ then $p^k = a^2+b^2$ in a unique way, then write $M= 2^c prod_{j=1}^J p_j prod_{i=1}^I q_i^{2e_i+d_i} $ where $p_j,q_i$ are primes $equiv 1$ and $3 bmod 4$ and $d_i in {0,1}$
$endgroup$
– reuns
Jan 9 at 20:27
$begingroup$
Maybe it will get the attention it deserves if it were posted on math overflow.
$endgroup$
– user25406
Jan 16 at 15:40
add a comment |
$begingroup$
Here :
Can the sum of two squares be used to factor large numbers?
an idea to factor a large number $N=a^2+b^2$ is shown.
Under which conditions does a representation $$N^2=u^2+v^2$$ exist , such that $gcd(N,gcd(u,v))$ is a non-trivial factor of $N$ ?
I guess this is the case, when $N$ has at least one prime factor of the form $4k+1$ and is not a prime power, but I am not sure whether this is true and how it can be proven.
number-theory elementary-number-theory prime-numbers square-numbers
asked Jan 9 at 15:49
PeterPeter
48.1k1139133
$endgroup$
Here :
Can the sum of two squares be used to factor large numbers?
an idea to factor a large number $N=a^2+b^2$ is shown.
Under which conditions does a representation $$N^2=u^2+v^2$$ exist , such that $gcd(N,gcd(u,v))$ is a non-trivial factor of $N$ ?
I guess this is the case, when $N$ has at least one prime factor of the form $4k+1$ and is not a prime power, but I am not sure whether this is true and how it can be proven.
number-theory elementary-number-theory prime-numbers square-numbers
number-theory elementary-number-theory prime-numbers square-numbers
asked Jan 9 at 15:49
PeterPeter
48.1k1139133
asked Jan 9 at 15:49
PeterPeter
48.1k1139133
asked Jan 9 at 15:49
PeterPeter
48.1k1139133
asked Jan 9 at 15:49
PeterPeter
48.1k1139133
asked Jan 9 at 15:49
PeterPeter
48.1k1139133
48.1k1139133
1
$begingroup$
To enumerate all the ways such that $M = u^2+v^2$ It suffices to know that for $k$ the order of $p bmod 4$ then $p^k = a^2+b^2$ in a unique way, then write $M= 2^c prod_{j=1}^J p_j prod_{i=1}^I q_i^{2e_i+d_i} $ where $p_j,q_i$ are primes $equiv 1$ and $3 bmod 4$ and $d_i in {0,1}$
$endgroup$
– reuns
Jan 9 at 20:27
$begingroup$
Maybe it will get the attention it deserves if it were posted on math overflow.
$endgroup$
– user25406
Jan 16 at 15:40
add a comment |
1
$begingroup$
To enumerate all the ways such that $M = u^2+v^2$ It suffices to know that for $k$ the order of $p bmod 4$ then $p^k = a^2+b^2$ in a unique way, then write $M= 2^c prod_{j=1}^J p_j prod_{i=1}^I q_i^{2e_i+d_i} $ where $p_j,q_i$ are primes $equiv 1$ and $3 bmod 4$ and $d_i in {0,1}$
$endgroup$
– reuns
Jan 9 at 20:27
$begingroup$
Maybe it will get the attention it deserves if it were posted on math overflow.
$endgroup$
– user25406
Jan 16 at 15:40
1
1
$begingroup$
To enumerate all the ways such that $M = u^2+v^2$ It suffices to know that for $k$ the order of $p bmod 4$ then $p^k = a^2+b^2$ in a unique way, then write $M= 2^c prod_{j=1}^J p_j prod_{i=1}^I q_i^{2e_i+d_i} $ where $p_j,q_i$ are primes $equiv 1$ and $3 bmod 4$ and $d_i in {0,1}$
$endgroup$
– reuns
Jan 9 at 20:27
$begingroup$
To enumerate all the ways such that $M = u^2+v^2$ It suffices to know that for $k$ the order of $p bmod 4$ then $p^k = a^2+b^2$ in a unique way, then write $M= 2^c prod_{j=1}^J p_j prod_{i=1}^I q_i^{2e_i+d_i} $ where $p_j,q_i$ are primes $equiv 1$ and $3 bmod 4$ and $d_i in {0,1}$
$endgroup$
– reuns
Jan 9 at 20:27
$begingroup$
Maybe it will get the attention it deserves if it were posted on math overflow.
$endgroup$
– user25406
Jan 16 at 15:40
$begingroup$
Maybe it will get the attention it deserves if it were posted on math overflow.
$endgroup$
– user25406
Jan 16 at 15:40
add a comment |
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1
$begingroup$
To enumerate all the ways such that $M = u^2+v^2$ It suffices to know that for $k$ the order of $p bmod 4$ then $p^k = a^2+b^2$ in a unique way, then write $M= 2^c prod_{j=1}^J p_j prod_{i=1}^I q_i^{2e_i+d_i} $ where $p_j,q_i$ are primes $equiv 1$ and $3 bmod 4$ and $d_i in {0,1}$
$endgroup$
– reuns
Jan 9 at 20:27
$begingroup$
Maybe it will get the attention it deserves if it were posted on math overflow.
$endgroup$
– user25406
Jan 16 at 15:40