Ring of rational integers is not uniquely Euclidean for any norm

Statement
The ring of rational integers cannot be made a uniquely Euclidean domain with any norm.

Proof
The proof can be broken down into the following steps:


 * 1) We cannot have $$a \ne 0$$ such that $$N(a) < N(1)$$: If such an $$a$$ exists, we have $$1 = 1.(1-a) + a$$ and also $$1 = 1.1 + 0$$: two Euclidean divisions for $$1$$ by itself.
 * 2) We cannot have $$a \ne 0$$ such that $$N(a) < N(-1)$$: If such an $$a$$ exists, we have $$-1 = -1.(1+a) + a$$ and also $$-1 = -1.1 + 0$$: two Euclidean divisions for $$-1$$ by itself.
 * 3) Thus, $$N(1) = N(-1)$$ and no element has smaller norm.
 * 4) Suppose $$a \ne 0, -1$$. Then at most one of these holds: $$N(a) < N(a+1)$$ or $$N(-1) < N(a + 1)$$. This follows from the uniqueness of Euclidean division for $$a$$ by $$a + 1$$.
 * 5) $$N(a) = N(1)$$ for all positive integers $$a$$: We prove this by induction on $$a$$. Clearly, $$N(1) =  N(1)$$. Suppose $$N(a) = N(1)$$.By the previous step, either $$N(a) < N(a+1)$$ or $$N(-1) <  N(a+1)$$. But by the induction hypothesis, $$N(a) = N(1)$$, which in turn is equal to $$N(-1)$$, so $$N(a) < N(a + 1) \iff N(-1) < N(a + 1)$$. Thus, neither of these holds, so $$N(a+1) \le N(a) = N(1)$$. Step (3) then forces $$N(a+1) = N(1)$$.
 * 6) $$N(-b) = N(1)$$ for all positive integers $$b$$: The proof is analogous to the preceding one, relying on the division of $$-b$$ by $$-b-1$$.
 * 7) Combining the last two steps, all elements have equal norm, and this is clearly not a Euclidean norm.