$2=(a+b\sqrt{10})(c+d\sqrt{10})\implies 2=(a-b\sqrt{10})(c-d\sqrt{10})$ 
Why, $2=(a+b\sqrt{10})(c+d\sqrt{10})\implies 2=(a-b\sqrt{10})(c-d\sqrt{10})$

To show is, that $2$ is irreducible in $\mathbb Z[\sqrt{10}]$ but not prime, everything is clear except the line above. So if the implication above is true, then I get an equation in $\mathbb Z$, $4=(a^2-10b^2)(c^2-10d^2)$ which has no nontrivial factorization, hence irreducibility is shown, BUT why $b$ and $d$ can change their signs ?

 A: You can work this out longhand.
Multiply the first expression out so that $$2=ac+10bd+(ad+bc)\sqrt {10}$$
Now this sets a rational number $2-ac-10bd$ equal to a multiple of $\sqrt {10}$. Since $\sqrt {10}$ is irrational, this multiple must be zero, so that $ad+bc=0$. And if this is true, you will find that the second expression comes to the same thing.
It is worth doing longhand to understand what Henning Makholm says in his solution (which is part of a more general theory)
A: All quadratic number fields/rings come with a conjugation: $a+b\sqrt D \mapsto a-b\sqrt D$. This is clearly a homomorphism, so just apply that to the first equation.
A: The point is that $\{1, \sqrt{10} \}$ is linearly independent over $\mathbb{Q}$, for otherwise, we have $x + y \sqrt{10} =0$ for some $x,y \in \mathbb{Q}$ not both $0$, which implies that $\sqrt{10} \in \mathbb{Q}$, which is not the case ( by uniqueness of prime factorization in $\mathbb{Z}$, for example).   
Now we have if we have $2 = (a+b \sqrt{10})(c+d\sqrt{10}) $ for $a,b,c,d \in \mathbb{Z}$
then the linear independence of $\{1,\sqrt{10} \}$ over $\mathbb{Q}$ forces $2 = ac +10bd$ and $ 0 = (ad+bc)$. But substituting this back also gives $2 = (a-b \sqrt{10})(c-d\sqrt{10})$.
This type of argument works with $\sqrt{10}$ replaced by $\sqrt{d}$ as long as $d$ is an integer which is not a perfect square.
