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)
 A: 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*}
A: Consider the expansion of $(a+b\sqrt c)^n$ starting from $a^n$. After that term we have $[k_1a^{n-1}(b\sqrt c)]$, $k_2a^{n-2}(b\sqrt c)^2$, $[k_3a^{n-2}(b\sqrt c)^3]$, $k_4a^{n-3}(b\sqrt c)^4$ and so on until $(b\sqrt c)^n$ (where the $k_i$ are binomial coefficients, which are irrelevant for this argument). The bracketed terms contain an odd power of $\sqrt c$, and thus reduce to some integer multiple of $\sqrt c$; their sum is therefore $e\sqrt c$. Likewise, the non-bracketed terms reduce to integers, and sum to d.
Now replace $\sqrt c$ with $-\sqrt c$. The bracketed terms are negated because they have an odd power of $-b\sqrt c=-1\times b\sqrt c$ and $(-1)^n=-1$ for any odd n. ($(-k)^n=(-1)^nk^n$ for all k.) Hence the $e\sqrt c$ becomes $-e\sqrt c$. The non-bracketed terms still sum to $d$ because they have even powers of $-b\sqrt c$ and $(-1)^n=+1$ for even n.
So the statement in the original question is true, given that c is not a perfect square.
A: Since $\sqrt{c}$ is irrational, then
$$
(a+b\sqrt{c})^n=\sum_{i=0}^n \binom{n}{i}a^{n-i}b^ic^{i/2}=\underbrace{\sum_{i \text{even}}\binom{n}{i}a^{n-i}b^ic^{i/2}}_{d}+\sqrt{c}\underbrace{\sum_{i\text{ odd}}\binom{n}{i}a^{n-i}b^ic^{(i-1)/2}}_{e}.
$$
On the other hand
$$
(a-b\sqrt{c})^n=\sum_{i \text{even}}\binom{n}{i}a^{n-i}(-b)^ic^{i/2}+\sqrt{c}\sum_{i\text{ odd}}\binom{n}{i}a^{n-i}(-b)^ic^{(i-1)/2}=d-e\sqrt{c}.
$$
