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Show that $3+\sqrt{-7}$ is an irreducible element in the integral domain $Z[-7]$

My approach. Equation 1: $$ 3+\sqrt{-7} = (a+b\sqrt{-7} )(c+d\sqrt{-7})$$ Equation 2: $$ 3-\sqrt{-7} = (a-b\sqrt{-7} )(c-d\sqrt{-7})$$ Multiplying above equations $$ 16 = (a^2+7b^2)(c^2+7d^2)$$ Possible solutions of the above equation in this ring are $$ 16 = 16*1 $$ $$ 16 = 1*16 $$ $$ 16 = 2*8 $$ $$ 16 = 8*2 $$ $$ 16 = 4*4 $$ First & second solutions would make one of the elements as unit. Hence ignoring them. 3rd and fourth solutions are not possible. Fifth solution is possible when $ a=2,c=2,b=0 $ and $ d=0$. Now I have two questions.

  1. fifth solution is not satisfying the first equation. why?
  2. How come the term $3+\sqrt{-7}$ is irreducible if we can express it as a product of two non-unit terms.
Anuj
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    You get $a=\pm 2$ and $c=\pm 2$ and if you put these back in the original equations you will see that they don't provide factors. – Mark Bennet Oct 23 '18 at 14:53
  • You can compare your ideas with many similar questions and answers here about this topic, e.g., here etc. – Dietrich Burde Oct 23 '18 at 15:21
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    Your inferences are not bidrectional, they only prove the $(\Rightarrow)$ direction, i.e. if such a factorization exists then $a,b$ satisfy said equations. The reverse direction needn't hold, i.e. solutions for $a,b$ needn't yield a factorization, i.e. the solutions $,a,b,$ may be "extraneous". So you need to verify that those possible solutions are actual solutions, i.e. you need to also (dis)prove the reverse direction $(\Leftarrow)$ – Bill Dubuque Oct 23 '18 at 15:45
  • As for $(2)$ you didn't actually factor it since you didn't verify that the possible solutions are actual solutions (they aren't). – Bill Dubuque Oct 23 '18 at 15:51

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