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$(\sqrt{a}+b)^n=N+f$ where $f \in (0,1)$

$(\sqrt{a}+b)^{n+2} =M+g$ where $g \in (0,1)$

Given that $0<\sqrt{a}-b<1$ and $(a,b)$ belongs to integers, then

  1. If $n$ is odd, $f>g$

  2. If $n$ is even, $f<g$

How to prove/disprove it?

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I am new to StackExchange,could anyone edit my post to a readable form): – Tom Lynd Sep 10 '13 at 16:50
Please see here for a guide to writing math with MathJax. Have I interpreted your question correctly with my edit? – Zev Chonoles Sep 11 '13 at 4:59
yeah,thanks a million:) – Tom Lynd Sep 11 '13 at 5:00
What is your exact definition of surd? Any restrictions on $a,b,n$ other than $0<\sqrt{a}-b<1$ and trivially $n>0,b\ne0$? Otherwise both parts are wrong (and your proof for part 1 must have a gap)! With obvious indexing of $f,g$ you get for $a=\frac{1}{2}$ and $b=\frac{1}{2}$ for odd $n$ $$f_1 \approx 0.2071 < g_1=f_3\approx0 .7589$$ and for even $n$ $$f_2\approx .4571 > g_2=f_4\approx .1231$$ – gammatester Sep 11 '13 at 8:03
A third and IMO better alternative: You can answer your own question and show the proof(s). This gives reputation and will be informative to the other users, who up-voted your question or marked it as favourite. – gammatester Sep 17 '13 at 8:46
up vote 1 down vote accepted

Since $a$ and $b$ are integers (and $a > 0$ is not a square, otherwise $0 < \sqrt{a}-b < 1$ would not hold), we have

$$\begin{align} (b+\sqrt{a})^n + (b-\sqrt{a})^n &= \sum_{k=0}^n \binom{n}{k}b^{n-k}a^{k/2} + \sum_{k=0}^n (-1)^k\binom{n}{k}b^{n-k}a^{k/2}\\ &= \sum_{k=0}^n \bigl(1 + (-1)^k\bigr)\binom{n}{k}b^{n-k}a^{k/2}\\ &= 2\sum_{m=0}^{\lfloor n/2\rfloor} \binom{n}{2m}b^{n-2m}a^m, \end{align}$$

so $(b+\sqrt{a})^n + (b-\sqrt{a})^n$ is an integer. From $0 < \sqrt{a}-b < 1$ it follows that $0 < \lvert b-\sqrt{a}\rvert^n < 1$ and $(b-\sqrt{a})^n$ is positive when $n$ is even, and negative when $n$ is odd.

Hence for odd $n$, we have

$$(\sqrt{a}+b)^n = \underbrace{(b+\sqrt{a})^n + (b-\sqrt{a})^n}_{N(n)} + \underbrace{(\sqrt{a}-b)^n}_{f(n)}$$

and with the notations of the problem, $g = f(n+2) = (\sqrt{a}-b)^2\cdot f(n) = (\sqrt{a}-b)^2\cdot f < f$.

For even $n$ we have

$$(\sqrt{a}+b)^n = \underbrace{(b+\sqrt{a})^n + (b-\sqrt{a})^n - 1}_{N(n)} + \underbrace{1-(\sqrt{a}-b)^n}_{f(n)},$$

and in the notation of the problem $g = 1-(\sqrt{a}-b)^{n+2} > 1 - (\sqrt{a}-b)^n = f$.

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