number field that is not norm-Euclidean

Proposition.  The real quadratic field $\mathbb{Q}(\sqrt{14})$ is not norm-Euclidean.

Proof.  We take the number  $\gamma =\frac{1}{2}+\frac{1}{2}\sqrt{14}$  which is not integer of the field ($14\equiv 2(mod4)$).  Antithesis:  $\gamma =\varkappa +\delta$  where  $\varkappa =a+b\sqrt{14}$  is an integer of the field ($a,b\in \mathbb{Z}$) and

Thus we would have

And since  ${(2a-1)}^2=4(a-1)a+1\equiv 1(mod8)$,  it follows  $E\equiv 1-14\cdot 1\equiv 3(mod8)$,  i.e.  $E=3$.  So we must have

${(2a-1)}^2\equiv {(2a-1)}^2-14{(2b-1)}^2\equiv 3(mod7).$(1)

But  $\{0,\pm 1,\pm 2,\pm 3\}$  is a complete residue system modulo 7, giving  the set  $\{1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}4}\}$  of possible quadratic residues modulo 7.  Therefore (1) is impossible.  The antithesis is wrong, whence the theorem 1 of the parent entry (http://planetmath.org/EuclideanNumberField) says that the number field is not norm-Euclidean.

Note.  The function N used in the proof is the usual

$$\text{N}:r+s\sqrt{14}\mapsto r^2-14s^2\mathit{\hspace{1em}}(r,s\in \mathbb{Q})$$

defined in the field $\mathbb{Q}(\sqrt{14})$.  The notion of norm-Euclidean number field is based on the norm (http://planetmath.org/NormAndTraceOfAlgebraicNumber).  There exists a fainter function, the so-called Euclidean valuation, which can be defined in the maximal orders of some algebraic number fields (http://planetmath.org/NumberField); such a maximal order, i.e. the ring of integers of the number field, is then a Euclidean domain.  The existence of a Euclidean valuation guarantees that the maximal order is a UFD and thus a PID.  Recently it has been shown the existence of the Euclidean domain $\mathbb{Z}[\frac{1+\sqrt{69}}{2}]$ in the field $\mathbb{Q}(\sqrt{69})$ but the field is not norm-Euclidean.

The maximal order $\mathbb{Z}[\sqrt{14}]$ of $\mathbb{Q}(\sqrt{14})$ has also been proven to be a Euclidean domain (Malcolm Harper 2004 in Canadian Journal of Mathematics).