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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.

Peter
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    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 $d_i \in {0,1}$ – reuns Jan 09 '19 at 20:27
  • Maybe it will get the attention it deserves if it were posted on math overflow. – user25406 Jan 16 '19 at 15:40

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