5
$\begingroup$

Given complex number $C=a+ib$, I want to find two complex numbers $C_1=x+iy$ and $C_2=z+iw$ such that $C=C_1*C_2$ (a,b,x,y, z and w are all non zero integers). This problem is at least as hard as Integer factoring. Prime complex number has one as its only factor.

Does this problem reduce to integer factoring? Is it NP-hard?

$\endgroup$
  • $\begingroup$ Such complex numbers are known as Gaussian integers. $\endgroup$ – Juho Sep 5 '13 at 22:30
  • $\begingroup$ Not quite. Gaussian integers can have zero real or imaginary parts. For example, 2 and -4i are Gaussian integers. But OP is requiring the real and imaginary parts to be nonzero integers. $\endgroup$ – JeffE Sep 7 '13 at 17:06
4
$\begingroup$

If $C = \prod_i C_i$ is a prime factorization of $C$ then $N(C) = \prod_i N(C_i)$, where $N(\alpha + \beta i) = \alpha^2 + \beta^2$. Furthermore, $\pi$ is prime if either (i) $N(\pi) = 2$, (ii) $N(\pi) = 4a+1$ is prime, (iii) $N(\pi) = (4a+3)^2$ is the square of a prime. So factoring $C$ reduces to factoring $N(C)$.

$\endgroup$
  • $\begingroup$ Thanks Yuval. Could you give the details of the derivations of those statements? Are you taking into account the phase of complex numbers during factorization? $\endgroup$ – Mohammad Al-Turkistany Sep 6 '13 at 10:57
  • 2
    $\begingroup$ Look up any standard account of Gaussian integers. $\endgroup$ – Yuval Filmus Sep 6 '13 at 12:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.