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Wouldn't it be because in order for ANY number to evenly divide into a sum, it must divide into each term in the sum evenly?Why does it follow that if x is an integer such that sqrt(-2)|x+sqrt(-2) in Z[sqrt(-2)], then

sqrt(-2)|x?

- Feb 5, 2012

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Hi Poirot,Why does it follow that if x is an integer such that sqrt(-2)|x+sqrt(-2) in Z[sqrt(-2)], then

sqrt(-2)|x?

Can you please elaborate more about your question. Do you mean,

\[\frac{x}{\sqrt{-2}}+\sqrt{-2}\in \mathbb{Z}[\sqrt{-2}]=\{a+b\sqrt{-2} ; a\in \mathbb Z,b\in \mathbb Z \}\]

then prove that, \(\sqrt{-2}|x\) ?

I believe the OP is asking that if \(\displaystyle \displaystyle \sqrt{-2}\) evenly divides \(\displaystyle \displaystyle x + \sqrt{-2}\), then to show that \(\displaystyle \displaystyle \sqrt{-2}\) evenly divides x.Hi Poirot,

Can you please elaborate more about your question. Do you mean,

\[\frac{x}{\sqrt{-2}}+\sqrt{-2}\in \mathbb{Z}[\sqrt{-2}]=\{a+b\sqrt{-2} ; a\in \mathbb Z,b\in \mathbb Z \}\]

then prove that, \(\sqrt{-2}|x\) ?

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- Jan 26, 2012

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I don't follow. We have $5|2+3$, yet five divides neither of the summands.Wouldn't it be because in order for ANY number to evenly divide into a sum, it must divide into each term in the sum evenly?

Are you referencing a certain property of this ring?

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To be more precise, if x|y+z and x|y, then x|z.I don't follow. We have $5|2+3$, yet five divides neither of the summands.

Are you referencing a certain property of this ring?

Pf. xm=y+z and xn=y for some m,n. Thus z=x(m-n) so that x|z

- Jan 26, 2012

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Agreed![HR][/HR]

To be more precise, if x|y+z and x|y, then x|z.

Pf. xm=y+z and xn=y for some m,n. Thus z=x(m-n) so that x|z