The ring $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$ is a UFD. Contradiction? Considering the ring $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$, I know that this ring is an euclidian domain and therefore a unique factorization domain.
Now, $23=(3+4\sqrt{2})(3-4\sqrt{2})=(11+7\sqrt{2})(11-7\sqrt{2})$. 
$(3+4\sqrt{2}),(3-4\sqrt{2}),(11+7\sqrt{2}),(11-7\sqrt{2})$ are irreducibles in $\mathbb{Z}[\sqrt{2}]$, since for example: $$N(3+4\sqrt{2})=(3+4\sqrt{2})(3-4\sqrt{2})=-23$$Where $N$ is the norm $N(a+b\sqrt{d})=(a+b\sqrt{d})(a-b\sqrt{d})$. Since $23$ is prime, $(3+4\sqrt{2})$ is irreducible.
Therefore $23$ is not expressed by a unique product of irreducibles, doesn't this contradicts the fact that $\mathbb{Z}[\sqrt{2}]$ is UFD?

There is just a calculation mistake, since 
$$
(3+4\sqrt{2})(3-4\sqrt{2}) = 9 - 32 = -23
$$
