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Show that there is a disc in $\mathbb{C}$ with radius R , so that no primes of $\mathbb{Z}[i]$ are contained in the disc.

I was thinking of taking a disc which which does not touch (0,0), for example : $|z-R|+ R<|R|$

But then how does one show that this disc doesn't contain any primes in $\mathbb{Z}[i]= \mathbb{Z}+\mathbb{Z}i$.

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    @VVV: You might want to thank Thomas Andrews; but you still need to say what $R$ is. Is it arbitrary? If so, try something like the argument used to show that every for every $d\gt 0$ there are $d$ consecutive integers, none of which are primes.2011-11-01

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It suffices to show for $R$ a positive integer.

Let $P$ be the product of all $z\in \mathbb{Z}[i]$ with $0<|z|<4R$.

Let $D = P + 2R$.

Then if $|z-D|, then $R<|z-P|<3R$. So $z-P$ is a divisor of $P$.

But then $z-P$ divides $z=(z-P) + P$.

On the other hand, since $R<|z-P|$, $z-P$ is not a unit of $\mathbb{Z}[i]$, so $z$ is not a prime.

You have to show additionally that $u(z-P)\neq z$ is not possible for any unit $u\in \mathbb{Z}[i]$, but that's not hard.

As noted above, $|z-P|<4R$.

On the other hand $|z| + R \geq |z| + |z-D| \geq |D| \geq |P| - 2R$. So $|z|\geq |P|-3R$.

Since it is easy to show that $|P|\geq 81R^4>7R$, we see that $|z|>4R$. So $z$ is not a unit times $z-P$.

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    Which is similar to the case of $\mathbb Z$, where, to find $n$ consecutive non-primes, we choose $(n+1)!+2,...,(n+1)!+(n+1)$. We have to avoid $(n+1)!\pm 1$, because it isn't obvious that these are not prime.2011-11-01