$(5 + \sqrt{2})(2-\sqrt{2})=(11-7\sqrt{2})(2+\sqrt{2})$ - Unique factorization? Personal question : We know that $5 + \sqrt{2}$, $2-\sqrt{2}$, $11-7\sqrt{2}$ and $2+\sqrt{2}$ are irreductible in $\mathbb{Z}[\sqrt{2}]$ and that $$(5 + \sqrt{2})(2-\sqrt{2})=(11-7\sqrt{2})(2+\sqrt{2}).$$ Why this fact doesn't contradict the unique factorization in $\mathbb{Z}[\sqrt{2}]$?
Is it because $(5 + \sqrt{2})(2-\sqrt{2})=(5 + \sqrt{2})(2+\sqrt{2})(3-2\sqrt{2})=(11 -7\sqrt{2})(2+\sqrt{2})$?
Is anyone could give me a full explication in ''Answer the question''? 

This is because the factorisation is unique up to units, and $\Bbb Z[\sqrt 2]$ has units, for example $(\sqrt 2+1)$.
In $\Bbb Z$, you had $15 = 3 \times 5 = (-3) \times (-5)$. This doesn't contradict the unique factorisation theorem either because each factor of one factorisation differs from a factor in the other by a unit (here, $-1$)
In your case you can tell by looking at the norms, that $(5+\sqrt 2)$ and $(11-7\sqrt 2)$ may be associates, and we can easily check this :
$(5+\sqrt 2)/(11-7\sqrt 2) = (5+\sqrt 2)(11+7\sqrt 2)/23 = (69+46\sqrt 2)/23 = 3+2\sqrt 2$.
Meanwhile, $(2-\sqrt 2)/(2+\sqrt 2) = (2-\sqrt 2)^2/2 = (6-4\sqrt 2)/2 = 3-2\sqrt 2$.
And we have $(3+2\sqrt 2)(3-2\sqrt 2) = 1$ so really, you go from one factorisation to the other by taking a $(3\pm 2\sqrt 2)$ factor from one irreducible and giving it to the other :
$(5+2\sqrt 2)(2-\sqrt 2) = (5+2\sqrt 2)(3-2\sqrt 2)(2+\sqrt 2) = (11-7\sqrt 2)(2+\sqrt 2)$
In $\Bbb Z$ there are only $2$ units, so you can just decide to only use positive representatives of primes and put a $(-1)$ to the side, and then you can immediately tell if two factorisations are different.
In $\Bbb Z[\sqrt 2]$ there isn't really a nice choice of representatives to be made.
