Prove 29 is not irreducible in Z

Firepanda · Apr 20, 2010

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

CaffeineJunky · Apr 20, 2010

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

Hurkyl · Apr 20, 2010

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.

Prove 29 is not irreducible in Z

Discussion

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