Given that "$n$" belongs to prime numbers and is greater than 2 ; $(a,b)$ belongs to integers and $0<\sqrt{a}-b<1$
$(\sqrt{a}+b)^n=N+f$ where $f \in (0,1)$
Show that $N$-$2b^n$ is divisible by $2abn$
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2$a = 2, , b = 1,, n = 2$: $(\sqrt{2}-1)^2 = 3+2\sqrt{2} = 5 + 2(\sqrt{2}-1)$. $5 - 2b^2 = 3$ is not divisible by $2abn = 8$. – Daniel Fischer Sep 11 '13 at 11:43
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Duh, that should have been a $+$, $(\sqrt{2}+1)^2 = 3+2\sqrt{2}$. – Daniel Fischer Sep 11 '13 at 11:50
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For $ n $ is odd when $ n > 2 $, $ N = (\sqrt{a}+b)^n - (\sqrt{a}-b)^n $. Expanding it we can get: $$ C_{n}^{2k} (\sqrt{a})^{2k}(b)^{n-2k} - C_{n}^{2k} (\sqrt{a})^{2k}(-b)^{n-2k} = 2C_{n}^{2k} (a)^{2k}(b)^{n-2k} $$ $$ C_{n}^{2k-1} (\sqrt{a})^{2k-1}(b)^{n-2k+1} - C_{n}^{2k-1} (\sqrt{a})^{2k-1}(-b)^{n-2k+1} = 0 $$
easymath3
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@ easymath3 There's a jump from step 2 to 3,can you explain it?By the way,when n is not a prime number,just an odd then N-2b^n would not always be divisible by n – Tom Lynd Sep 11 '13 at 14:02
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@Tom Lynd YES, you must prove that $ n | C_{n}^{2k} $, when n is prime it's obvious. – easymath3 Sep 11 '13 at 14:12
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