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Divisibility and Perfect Power: \(2\le n,a,b,k=(a^n+b^n)/((ab)^{n-1}+1) \in \mathbb{N} \Rightarrow \sqrt[n]{k} \in \) Z
05-17-2010, 09:31 AM
Post: #1
Divisibility and Perfect Power: \(2\le n,a,b,k=(a^n+b^n)/((ab)^{n-1}+1) \in \mathbb{N} \Rightarrow \sqrt[n]{k} \in \) Z
Proof We have \( a = b, \; k=\displaystyle{\frac{a^n+b^n}{(ab)^{n-1}+1}} \in \mathbb{N}^+ \Rightarrow a=b=k=1=1^n\)
Now assume that \(a<b,\;n>1,\; a^n-k=(ka^{n-1}-b)b^{n-1}\)
(1) If \(k > a^n\) then \(k > k-a^n = b^{n-1} (b-ka^{n-1}) \ge b^{n-1}\)
\(\quad \;\) thus \(b > ka^{n-1} \ge k > b^{n-1}\). It's impossible.
(2) If \(k < a^n\), then \(ka^{n-1}-b = \displaystyle{\frac{a^n-k}{b^{n-1}}} < a\), \(ka^{n-1} < a+b\), \((k-1)a^{n-1} < a+b - a^{n-1} \le b\)
\(\quad\) If \(k > 1\), then \(a^{n-1} \le (k-1)a^{n-1} < b\), \( a^n-k = (ka^{n-1}-b)b^{n-1} \ge b^{n-1} > (a^{n-1})^{n-1} \ge a^n\),
\(\quad \;\)that's wrong and so \(k=1=1^n\)
(3) If \(k = a^n\) then \(b=ka^{n-1}=a^{2n-1} \; b^n = a^{2n^2-n} = a^{2n(n-1) + n}\) hence \(\displaystyle{\frac{a^n+b^n}{(ab)^{n-1}+1}=\frac{a^n(1+a^{2n(n-1)})}{(a^{2n})^{n-1}+1}} = a^n\)
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05-18-2010, 12:20 PM
Post: #2
Divisibility & Perfect Power: \(2\le n,a,b,k=(a^n+b^n)/((ab)^{n-1}+1) \in \mathbb{N} \Rightarrow \sqrt[n]{k} \in \) Z
This is the general case of http://math.elinkage.net/showthread.php?tid=67
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