Units in the cyclotomic ring of degree $7$ Let $\zeta=e^{2\pi i \over 7}$.  Given $u,v\in\mathbb{Z}[\zeta]$, show that:
$$ u\bar{u}+v\bar{v}\,\,\text{ is a  unit }\implies (1-\zeta)\mid uv. $$
Motivation An equivalent form of this result is often used to demonstrate various phenomena in low dimensional topology (see my comment on this question.  Note "$\zeta\mapsto -1$" is a typo and should be  "$\zeta\mapsto 1$").  As mentioned in the comment, authors usually cite this as a special case of part of a calculation buried deep in a $106$ page paper.
Attempted approach If one could find criteria for elements of the form $u\bar{u}\in\mathbb{Z}[\zeta+\zeta^{-1}]$, with $ u\in\mathbb{Z}[\zeta]$ and there was a nice description of the units in $\mathbb{Z}[\zeta+\zeta^{-1}]$, one might be able to obtain a constraint on expressing them as the sum of two elements satisfying the criterion. However $\zeta+\zeta^{-1}$ is a unit with infinite multiplicative order, and it is not clear what a nice description of the units might be.

Update A search through all "small" $u,v$ found that every time that $u\bar{u}+v\bar{v}$ is a unit, we have $uv=0$.  Here "small" means expressible as a linear combination of $1,\zeta,\zeta^2,\cdots,\zeta^6$ with coefficients differing by at most $5$.  It therefore seems reasonable to conjecture the stronger statement:$$ u\bar{u}+v\bar{v}\,\,\text{ is a  unit }\implies uv=0, $$ which may be easier to prove.
 A: This appears to be much simpler than I thought.
Suppose $u\bar{u}+v\bar{v}$ is a unit with $u,v\in \mathbb{Z}[\zeta]$ both non-zero.  We will draw a contradiction.
Let $*\colon \mathbb{Z}[\zeta] \to \mathbb{Z}[\zeta]$ be the automorphism mapping $\zeta\mapsto \zeta^2$.  Any $x\in \mathbb{Z}[\zeta]$ can be written as a linear combination of $1,\zeta,\cdots, \zeta^6$ with coefficients in $\mathbb{Z}$.  If $x=x^*$ and $x=\bar{x}$ then its coefficients on $\zeta,\cdots, \zeta^6$ must be equal and we have $x\in \mathbb{Z}$.
If $u\bar{u}+v\bar{v}$ is a unit then so are $(u\bar{u}+v\bar{v})^*$ and $(u\bar{u}+v\bar{v})^{**}$, so the product of all three:
$$(u\bar{u}+v\bar{v})(u\bar{u}+v\bar{v})^*(u\bar{u}+v\bar{v})^{**},$$
is a unit too.
The elements $$(u\bar{u}+v\bar{v})(u\bar{u}+v\bar{v})^*(u\bar{u}+v\bar{v})^{**},\qquad (u\bar{u})(u\bar{u})^*(u\bar{u})^{**},\qquad (v\bar{v})(v\bar{v})^*(v\bar{v})^{**},$$ are all integers, as they are invariant under $*$ and complex conjugation.  Further they are all positive integers, as they are each the product of three positive real numbers.  Thus $$(u\bar{u})(u\bar{u})^*(u\bar{u})^{**},\, (v\bar{v})(v\bar{v})^*(v\bar{v})^{**}\geq 1.\qquad\qquad(1)$$
The only positive integer unit is $1$, so $$(u\bar{u}+v\bar{v})(u\bar{u}+v\bar{v})^*(u\bar{u}+v\bar{v})^{**}=1.$$
Expanding we get: \begin{eqnarray*}1&=&(u\bar{u}+v\bar{v})(u\bar{u}+v\bar{v})^*(u\bar{u}+v\bar{v})^{**}\\
&=&(u\bar{u})(u\bar{u})^*(u\bar{u})^{**}\left(1+\frac{v\bar{v}}{u\bar{u}}+\frac{(v\bar{v})^*}{(u\bar{u})^*}+\frac{(v\bar{v})^{**}}{(u\bar{u})^{**}}\right)\\&+&(v\bar{v})(v\bar{v})^*(v\bar{v})^{**}\left(1+\frac{u\bar{u}}{v\bar{v}}+\frac{(u\bar{u})^{*}}{(v\bar{v})^{*}}+\frac{(u\bar{u})^{**}}{(v\bar{v})^{**}}\right)\\&\geq&1\cdot1+1\cdot 1=2.
\end{eqnarray*}
We have attained our desired contradiction $1\geq 2$.
