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I am thinking about the following question:

Let $phi= frac1+ sqrt 52$ be the golden ratio, and define:
$$ mathbb Z [phi]=a+ bphi : quad a, b in mathbb Z $$
We wish to prove that for $a,b in mathbb Z$ we have that $a+ bphi in mathbb Z[phi]^* iff a^2 +ab-b^2 = pm 1$.
Note: the star indicates the multiplicative group of our ring.

This exercise comes with a hint: let $barphi=frac1-sqrt 52$, You may use without proof that
$(a + b phi)(c + dphi) = 1$ if and only if $(a + bbarphi)(c + dbarphi) = 1$.

I have trouble interpreting how the hint plays out.

For $z = a + b phi$, we write $overlinez = a + b overlinephi$.

Define the function $N: mathbbZ[phi] rightarrow mathbbZ$ by $N(z)=z overlinez$. It's easy to check that $N(a+b phi) = a^2 +ab -b^2 in mathbbZ.$

Proposition: $z in mathbbZ[phi]$ is a unit if, and only if, $N(z) in mathbbZ$ is a unit.

Proof:

($Rightarrow$) Suppose $(a+bphi)(c+dphi)=1$. By the hint we have $(a+boverlinephi)(c+doverlinephi)=1$, and so $$N(a+bphi)N(c+dphi) = (a+bphi)(a+boverlinephi)(c+dphi)(c+doverlinephi) = 1.$$ Since $N(a+bphi)$, $N(c+dphi) in mathbbZ$, we must have either $$N(a+bphi) = N(c+dphi) = 1 textor N(a+bphi) = N(c+dphi) = -1.$$

($Leftarrow$) Suppose that $N(a+bphi) = (a+bphi)(a+boverlinephi) = (-1)^k, kin 0,1 $.

We want to find $c,d in mathbbZ$ such that $c+dphi = a+boverlinephi$. So write $$ c+dphi = c+ frac12 d + fracsqrt52 equiv a + frac12b -fracsqrt52 = a+boverlinephi,$$ and equate coefficients in $mathbbZ + sqrt5mathbbZ$ to get $c = a+b$ and $d=-b$. That is, we have $$(a+bphi)((-1)^k (a+b) + (-1)^k+1 bphi) = 1.$$

$hspace18cm square$

Therefore, given $a+bphi in mathbbZ[phi]$, $$a+bphi in mathbbZ[phi]^* Leftrightarrow a^2 +ab -b^2 = N(a+bphi ) = pm 1.$$

I do not see how the hint is of any use in answering the question. There are many different approaches, depending how much theory you are familiar with. I'll give a very basic approach, assuming very little theory.

Note that for $a,b,c,dinBbbZ$ we have $a+bphi=c+dphi$ if and only if $a=c$ and $b=d$. Also note that $phi+barphi=1$ and $phibarphi=-1$, meaning that both are roots of $X^2-X-1$. This shows that
$$BbbZ[phi] longrightarrow BbbZ[phi]: a+bphi longmapsto a+bbarphi,$$
is a ring automorphism of $BbbZ[phi]$, from which the hint follows. I do not see how the hint is relevant though.

We also see that for every $xinBbbZ[phi]$ there exist unique $a,b,a',b'inBbbZ$ such that
$$x=a+bphi=a'+b'barphi.$$
Now if $a,binBbbZ$ are such that $a^2+ab-b^2=pm1$ then
$$(a+bphi)(a+bbarphi)=a^2+ab(phi+barphi)+b^2phibarphi=a^2+ab-b^2=pm1,$$
which shows that $a+bphiinBbbZ[phi]^times$ with $(a+bphi)^-1=pm(a+bbarphi)$.

Conversely, if $a,binBbbZ$ are such that $a+bphiinBbbZ[phi]^times$ then there exist unique $c,dinBbbZ[phi]$ such that
$$(a+bphi)(c+dphi)=1.$$
Then $gcd(a,b)=1$, and there exist unique $c',d'inBbbZ$ such that
begineqnarray*
1&=&(a+bphi)(c+dphi)=(a+bphi)(c'+d'barphi)
=ac'+ad'barphi+bc'phi+bdphibarphi.
endeqnarray*
Because $phibarphi=-1$ and $phi+barphi=1$ it follows that
$$1=ac'+ad'barphi+bc'phi+bd'phibarphi=(ac'+ad'-bd')+(bc'-ad')phi,$$
which implies that $bc'=ad'$. Because $gcd(a,b)=1$ this shows that $c'=pm a$ and $d'=pm b$, where the two signs agree. Hence
$$1=(a+bphi)(c+dphi)=pm(a+bphi)(a+bbarphi)=pm(a^2+ab-b^2).$$
This proves that for all $a,binBbbZ$ we have
$$a+bphiinBbbZ[phi]^timesqquadiffqquad a^2+ab-b^2=pm1.$$