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Factor $65$ into irreducible in $\mathbb{Z}[i]$

I tried to factor $65$ in Gaussian integers by Mathematica, and I got $65 = -(1+2i)(2+i)(2+3i)(3+2i)$, but i don't know how to factor it by hand. Could you help me to solve this? Thank you!!!

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  • $\begingroup$ $65=8^2+1^2 = (8+i)(8-i)$. At least a start. $\endgroup$ – Teresa Lisbon Mar 24 '16 at 1:29
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It is enough to factor $5$ and $13$ separately. A factor of $5$ will a norm equal to a divisor of $N(5)=25$, i.e. if $a+bi$ divides $5$, $a^2+b^2$ divides $25$. The possible solutions are $1\pm2i$ and the multiple of these by one of the $4$ units in $\mathbf Z[i]$ ($1,-1,i,-i$).

Similarly, for $13$, we have to find Gauß' integers satisfying $a^2+b^2=13$. The basic solutions are $ 2\pm3i$. Thus you have a factorisation: $$65=u(1+2i)(1-2i)(2+3i)(2-3i),\quad u\in\{1,-1,i,-i\}$$ Check, one finds $u=1$.

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All irreducible nonunit elements $\alpha=a+bi$ of $ \mathbb Z[i]$ are :

  1. $\alpha$ is a prime number of the form $4m+3$
  2. $\alpha=1+i$ or $1-i$
  3. $N(\alpha)=a^2+b^2$ is a prime number of the form $4m+1$

Now $65=5\cdot13$ and $5,13$ is not of the forms 1,2 and $N(5)=25 , N(13)=169$ is not of the form 3, so they are must be reducible, $5=2^2+1$ and $13=3^2+2^2$ so $5=(1+2i)(1-2i)$ and $13=(3+2i)(3-2i)$ and all these factors are of the form 3.

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