$1-i$ is irreducible in $\mathbb{Z}[i]$ 
(1) How can I show that $1-i$ is irreducible over $\mathbb{Z}[i]$?

I tried that suppose $1-i$ is reducible over $\mathbb{Z}[i]$, then $$1-i=xy$$ where $x$ and $y$ are non units.
Norm is defined as: $N(a+b\sqrt{d})=|a^2-db^2|$.
$$N(1-i)=2=N(x)N(y)$$
I have two possibilities only:
$$N(x)=1 \ \ \text{or} \ \ N(y)=1$$
which is not possible. But I want know if there is another way to prove this.

(2) How can I show that $a^2-5b^2=2$ and $a^2+5b^2=23$ have no solution in $\mathbb{Z}$?

I'm confused in this type of questions. Can I find the contradiction without taking modulo system?
If there is no way then IS IT TRUE THAT SOME EQUATION HAS NO SOLUTION IN $\mathbb{Z}_p$ then IT HAS NO SOLUTION IN $\mathbb{Z}$? 
Please help me. Thanks in advance.
 A: (1) One can show that $1-i$ is prime (hence irreducible) in $\mathbb Z[i]$ as follows: 
$\mathbb Z[X]/(X^2+1)\simeq\mathbb Z[i]$ by sending $X$ to $i$. This way $1-i$ corresponds to (the residue class of) $1-X$, so $$\mathbb Z[i]/(1-i)\simeq\frac{\mathbb Z[X]/(X^2+1)}{(1-X,X^2+1)/(X^2+1)}\simeq\mathbb Z[X]/(1-X,X^2+1)\simeq\frac{\mathbb Z[X]/(1-X)}{(1-X,X^2+1)/(1-X)}\simeq\mathbb Z/2.$$ (In the last isomorphism we sent $X$ to $1$.) Since $\mathbb Z/2$ is an integral domain (actually a field) the ideal $(1-i)$ is prime, so $1-i$ is a prime element.
(2) I think your teacher and Jyrki's comments answered completely this question.
A: Let $N: \mathbb Z[i] \rightarrow \mathbb N$ denote the canonical norm function on $\mathbb Z[i]$ given by $N(z=a+ib) = z \bar z = a^2 + b^2$. It is easy to see that $N(xy) = N(x)N(y)$ for $x,y \in \mathbb Z[i]$.
In this question $z = 1-i \in \mathbb Z[i]$ and we must show $z$ is irreducible.
Suppose we have $xy= z$ then $N(xy) = N(x)N(y) = N(z) = 1^2 + 1^2 = 2$.
Since $2$ is a prime number we can only factor it as $1 \cdot 2 \lor 2 \cdot 1 \Rightarrow N(x) = 1 \lor N(y) = 1$. (1)
Now we must show $N(x) = 1$ if and only if $x \in \mathbb Z[i]^*$:
$x \in \mathbb Z[i]^* \Rightarrow xy = 1 \Rightarrow N(x)N(y) = N(1) = 1 \Rightarrow N(x) = 1$.
$N(x) = 1 \Rightarrow x \bar x = 1 \Rightarrow x^{-1} = \bar x \in \mathbb Z[i]$
This together with (1) implies $1-i$ can only be factored as a unit times a non-unit (norm is 2) and thus $1-i$ is irreducible by definition.
Often to show that elements are irreducible or units we use a norm function.
