If $z$ is a Gaussian integer such that $|z|^2$ is composite in $\mathbb{Z}$. Then is it always possible to express $z$ as the product of two Gaussian integers (The norm of all the factors are $>1$) ? I mean is there any Gaussian integer $z=a+ib$ such that $|z|^2=a²+b²$ is composite in $\mathbb{Z}$ but we can't get the from $a^2+b^2$ from the product of any two conjugate factors. $a^2+b^2$ is unique and self-sustained? In broad , suppose $z=a+ib$ is a gaussian integer . $|z|= a^2+b^2=(p^2+q^2)(m^2+n^2) . =(pm+qn)^2+(pn-qm)^2.....(1) . =(pn+qm)^2+(pm-qn)^2.....(2)$ Then for different factorization we would get different expressions. Can we always include $a^2+b^2$ (not the sum, but the expression) in the set of the forms like(1), (2)?
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The choice $z=3$ is the smallest counterexample. $|z|^2=9$ is composite in $\Bbb{Z}$, but $z$ cannot be written as a product of two Gaussian integers with norms $>1$. – Jyrki Lahtonen Apr 25 '19 at 07:51
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The generator $(a+bi)$ of $\mathbb{Z}[i]$ must have no further divisors or that divisor could itself act as a generator. So the factor $(a+bi)$ of composite $a^2+b^2$ cannot have a further factor, and this can be seen trivially, if $a+bi=(p+qi)(p-qi)$, note the R.H.S is purely real. Are you asking for this?
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I mean, if a gaussian integer z's norm is composite ,then can we certainly express z as a product of two gaussian integers. Suppose, z=a+ib, and c=a²+b²=(p²+q²)(m²+n²). I think it may happen that a²+b² is composite but we can not get a+ib =(p+qi)(m+in). – Alapan Das Apr 25 '19 at 04:21
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Yes I'm saying something like that. In short I'm speaking why all Gaussian integers with composite norm in Z is composite in Z[i] ? – Alapan Das Apr 25 '19 at 04:34
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1Yes. It's confusing. I can not write the question correctly. But I think the question is very deep. I shall try to find it . – Alapan Das Apr 25 '19 at 04:59
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