If $(a + b\sqrt c)^n = d + e\sqrt c$, then $(a - b\sqrt c)^n = d - e\sqrt c$ I think that:

If $a,b,(c\ge0$ not a prefect square$),d,e,f\in\mathbb Z$ such that for some $n \ge 1$, $(a + b\sqrt c)^n = d + e\sqrt c$, then $(a - b\sqrt c)^n = d - e\sqrt c$

Is this true? Can someone provide a proof or give a hint for how to prove this? (preferably without induction if it doesn't provide insight)

Consider the set
\begin{align*}
A = \{ x + y\sqrt{c}: x, y \in \mathbb{Z} \} 
\end{align*}
Consider the map $f: A \rightarrow A$ defined by $f(x+y\sqrt{c}) = x - y \sqrt{c}$. It is easy to see that $f$ is well defined (since $c$ is not a perfect square) and for any two $a_1, a_2 \in A$,
\begin{align*}
f(a_1+a_2) &= f(a_1) + f(a_2)\\
f(a_1 \cdot a_2) &= f(a_1)\cdot f(a_2)
\end{align*}
Thus if $(a + b\sqrt c)^n = d + e\sqrt c$, then 
\begin{align*}
f((a + b\sqrt c)^n) &= f(d + e\sqrt c) \\
(f(a+b\sqrt{c}))^n &= d-e\sqrt{c} \\
(a-b\sqrt{c})^n &= d-e\sqrt{c}
\end{align*}
