# Number TheoryZ[sqrt(-2)]

#### Poirot

##### Banned
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?

#### Prove It

##### Well-known member
MHB Math Helper
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?
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?

#### Sudharaka

##### Well-known member
MHB Math Helper
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?
Hi Poirot,

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

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

#### Prove It

##### Well-known member
MHB Math Helper
Hi Poirot,

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

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.

#### Poirot

##### Banned
Of course. Thanks 'Prove it'.

#### The Chaz

##### Member
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?
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?

#### Poirot

##### Banned
[HR][/HR]
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?
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

#### The Chaz

##### Member
[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
Agreed!