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In the ring of $\mathbb Z[i]$ use our division algorithm divide $7+2i$ by $2+i$.

$$\frac{7+2i}{2+i}\cdot \frac{2-i}{2-i}=\frac{16-3i}{5}=\frac{16}{5}-\frac{3}{5}i$$

Does this mean you cannot divide $7+2i$ by $2+i$, because the real and imaginary parts are $\frac{16}{5}$ and $\frac{3}{5}i$ which are not integers.

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Hmmm...perhaps it should be "divide with residue...etc."? After all, the Gaussian integers are an Euclidean Domain. –  DonAntonio Nov 16 '12 at 3:24
    
The quotient cannot be a Gaussian integer since (in the number-theoretic sense) $2+i$ has norm $5$ and $7+2i$ has norm $53$, which is not a multiple of $5$. –  André Nicolas Nov 16 '12 at 3:40
    
This reflects that in a ring you can't always divide. Rings are like the integers and you can't always divide there (without remainder). You need a field to be able to divide and, as the problem says, $\mathbb Z[i]$ is a ring. –  Ross Millikan Nov 16 '12 at 3:56

3 Answers 3

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Presumably, what's meant is "divide with remainder". So you have the quotient $(16/5)-(3/5)i$, which, rounding everything to the nearest integer, is $3-i$, with remainder $(1/5)+(2/5)i$. That is, $${7+2i\over2+i}=(3-i)+({1\over5}+{2\over5}i)$$ Multiply through by $2+i$ to get the nicer-looking $$7+2i=(3-i)(2+i)+i$$

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First you have $$ \mathbb{Z}[i] = \{a + bi: a,b\in \mathbb{Z}\}. $$ (Here $a$ and $b$ are unique.) So the question is: does there exist a $a+bi$ such that $$ (a+bi)(2+i) = 7 +2i? $$ So we would need $$ 2a - b + i(a + 2b) = 7 + 2i. $$ That is $$ \begin{align} 2a - b &= 7 \\ a + 2b &= 2. \end{align} $$ So you are basically asking if you can find $a,b\in \mathbb{Z}$ satisfying these two equations. As you have already realized, you can't. (You can find solutions in $\mathbb{Q}$, but not in $\mathbb{Z}$).

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Define $N(a+bi)=a^2+b^2$. Obviously, if $\alpha$ divides $\beta$ then $N(\alpha) $ divides $N(\beta)$. In this example $N(2+i)=5$ which does not divide $N(7+2i)=53$.

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