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 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.
