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Let $u$ and $v$ be two numbers of the form $u=a+b\sqrt{c}$, where a,b and c are rational numbers, with $\sqrt{c}$ an irrational constant. Let the notation $u^∗$ indicate the value of $u$ when the sign of b is reversed. That is, $u^∗ = a − b\sqrt{c}$.

Prove that $(u^n)^∗ = (u^∗)^n$

For the assumption when $n=k$, the solutions let $u^k = p+q\sqrt{c}$, and my question is why can the values of $a$ and $b$ be changed but not $c$? Additionally, it was let that $u=a+b\sqrt{c}$ and $v=p+q\sqrt{c}$ for the first parts of the question, to show that $$u^∗ + v^∗ = (u + v)^∗$$

Again, why can't I let $v = p + q\sqrt{m}$ instead?

user71207
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  • If you’re counting $\sqrt{c}$ in the expansion of $(a+b\sqrt{c})^n$ then always at the end you’ll find some terms with a $\sqrt{c}$ in them, and never find any terms with a square root of anything else (unless you pointlessly convert $4$ to $\sqrt{16}$, for example). Your question is very similar to some properties of the complex conjugate, if that prompts you – FShrike Jul 30 '21 at 08:41
  • Indeed it is a question from the complex number topic. But for the second part, the show question, it is still possible to prove by letting $v = p + q\sqrt{m}$ instead of $c$ – user71207 Jul 30 '21 at 08:49

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