I think 5 works, but I am not sure how to argue that 5 is prime in $\mathbb{Z}[\sqrt{2}]$. My thoughts were to show that when we mod by the ideal (5) that we get an integral domain, which I can't think of a quick way to argue this other than brute force (which is not an efficient use of time) or using that $\mathbb{Z}[\sqrt{2}]$ is a UFD so prime if and only if irreducible and using a field norm argument, in which case I get stuck trying to show that $a^{2} -2b^{2}=5$ has no integer solutions.

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    $\begingroup$ If you have $a^2 - 2b^2 \equiv 0 \pmod{5}$, can you see that that implies $a \equiv 0 \pmod{5}$? $\endgroup$ – Daniel Fischer May 15 '15 at 22:10
  • $\begingroup$ Larara, if $5 = rs$ in $\mathbb{Z}[\sqrt{2}]$ then $N(5) = N(r)N(s) = 25$, so $N(r)=N(s)=5$ or, WLOG, $N(r)=1$ and $N(s)=25$. If $r=a+b\sqrt{2}$ then $N(r)= a^{2} -2b^{2}$ which is what would be equal to $5$. $\endgroup$ – TuoTuo May 15 '15 at 22:17
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    $\begingroup$ Pedantically, one should mention the possibility of $N(r) = N(s) = -5$, but since $N(u) = -1$ has solutions in $\mathbb{Z}[\sqrt{2}]$, we can always achieve the case of positive norms. $\endgroup$ – Daniel Fischer May 15 '15 at 22:22
  • $\begingroup$ There is no element in $\alpha$ in $ \mathbb{Z[\sqrt{2}]}$ that have $N(\alpha) =5$. This is actually useful in proving that $5$ is a prime element $\endgroup$ – alkabary May 15 '15 at 22:40

You should know the Norm $N(\alpha) = a^2 - 2b^2$ where $\alpha = a + b\sqrt{2} \in \mathbb{Z[\sqrt{2}]}$ and so the norm of $5$ is $N(5) = 5^2 = 25$.

The Norm $N$ has many properties , the most important ones are listed below

(1) $N(\alpha \beta) = N(\alpha) N(\beta) \space \forall \space \alpha, \beta \in \mathbb{Z[\sqrt{2}]}$

(2) $N(\alpha) = 1 \iff \alpha$ is a unit in $\mathbb{Z[\sqrt{2}]}$

(3) if $\alpha \mid \beta$ in $ \mathbb{Z[\sqrt{2}]}$ then $N(\alpha) \mid N(\beta)$ in $ \mathbb{Z}$

Now you have all the tools to prove that $5$ is prime in $\mathbb{Z[\sqrt{2}]}$.

First of all you should know that the definition of a prime is as follows

An element $p$ is prime if and only if $p$ is not a unit and $p \mid ab \implies p \mid a$ or $p \mid b$.

so here is the proof that $5$ is prime in $ \mathbb{Z[\sqrt{2}]}$


$5$ is not a unit because $N(5) = 25$. Now assume $5 \mid \alpha\beta$ where $\alpha ,\beta \in \mathbb{Z[\sqrt{2}]}$. This implies by property (3) above that $N(5) \mid N(\alpha\beta)$ and which also implies by property (1) that $N(5) \mid N(\alpha)N(\beta)$ and so $$25 \mid N(\alpha)N(\beta)$$

Now it's easy to prove that no element in $ \mathbb{Z[\sqrt{2}]}$ has a Norm equal to $5$ because $a^2 - 2b^2 = 5 \implies a^2 = 5$ which implies that $a = \sqrt{5}$ but since we are dealing with $\mathbb{Z[\sqrt{2}]}$ then we don't have any elements with $N(\alpha) = 5$. This means that either $N(\alpha) = 25$ and $N(\beta) = 1$ or vice versa, In any case thus means that $\alpha = 5$ or $\beta = 5$ which means that $5 \mid \alpha$ or $5 \mid \beta$, Hence ,$5$ is a prime element in $\mathbb{Z[\sqrt{2}]}$.

As an exercise, I suggest that you proof these three Properties of the norm, and also try to find all the units in $ \mathbb{Z[\sqrt{2}]}$

To prove that there are no integers $a,b$ satisfying the equation $$a^2 -2b^2 = 5$$

Break it into cases.

(1) $-2b^2$ is always even and if $a$ is even then $a^2$ is also even and even - even = even and $5$ is odd so this will not work

(2) Now the only other option is to have $a$ is odd, try to figure out why it won't work either.

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  • $\begingroup$ I think you want $=$ rather than $\mid$ in your proof (i.e., $5=\alpha\beta$, etc.) $\endgroup$ – Steven Stadnicki May 15 '15 at 22:43
  • $\begingroup$ why ? I don't get it $\endgroup$ – alkabary May 15 '15 at 22:46
  • $\begingroup$ Because there are many $\alpha$ and $\beta$ such that $5|\alpha\beta$ - your argument isn't exactly wrong per se, but for instance you could take $\alpha=375\sqrt{2}$ and $\beta=11$. It's much easier to try saying $5=\alpha\beta$; the rest of your argument goes through intact, but then you do in fact have $N(\alpha)=25$ or $1$. $\endgroup$ – Steven Stadnicki May 16 '15 at 0:17
  • $\begingroup$ This is exactly what I was doing, except I don't comprehend how you made the jump that $a^2-2b^2 =5 \implies a^2=5$. $\endgroup$ – TuoTuo May 16 '15 at 3:15
  • $\begingroup$ well I am basically saying that $a^2 -2b^2 = 5$ doesn't have a solution were $a,b$ are both integers . $\endgroup$ – alkabary May 16 '15 at 3:16

Note that $$ R=\frac{\Bbb Z[\sqrt{2}]}{(5)}\simeq \frac{\Bbb Z[X]}{(5,X^2-2)}\simeq \frac{\Bbb F_5[X]}{(X^2-2)}. $$ Thus everything boils down to proving that $X^2-2$ is irreducible in $\Bbb F_5[X]$ and this is equivalent to $2$ not being a square modulo $5$. Indeed the squares in $\Bbb F_5=\Bbb Z/5\Bbb Z$ are $0$, $1=1^2=4^2$ and $4=2^2=3^2$.

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  • $\begingroup$ Thanks this is very clever also. $\endgroup$ – TuoTuo May 16 '15 at 3:46

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