Prove 29 is not irreducible in Z

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SUMMARY

29 is not irreducible in the ring of Gaussian integers Z[i] because it can be expressed as a product of two non-unit elements: (5 - 2i) and (5 + 2i). The proof demonstrates that (5 - 2i)(5 + 2i) equals 29, confirming that 29 can be factored in Z[i]. The discussion emphasizes the importance of recognizing that both factors are non-units in this context.

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Firepanda
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Prove 29 is not irreducible in Z

Don't I just have to show here that 29 is a product of 2 elemets in Z?

So

29 = (a+bi)(a-bi) where a=5 b=2

That can't just be it though?

Thanks
 
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Why wouldn't it be it?

Goal: Show that 29 can be factored over the Gaussian integers.
Proof (by explicit example): (5 - 2i)(5 + 2i) = 25 - 10i + 10i - 2i^2 = 25 + 4 = 29. QED
 


Firepanda said:
29 = (a+bi)(a-bi) where a=5 b=2

That can't just be it though?
Sure it is, depending on how obvious you consider it to be that 5+2i and 5-2i are non-units.
 

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