When can we factor $N$ efficiently with a representation $N^2=a^2+b^2$?












2












$begingroup$


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.










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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
















2












$begingroup$


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.










share|cite|improve this question









$endgroup$








  • 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














2












2








2





$begingroup$


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.










share|cite|improve this question









$endgroup$




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






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










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