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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
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
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 $$\displaystyle{\frac{a^n+b^n}{(ab)^{n-1}+1}=\frac{a^n(1+a^{2n(n-1)})}{(a^{2n})^{n-1}+1}} = a^n$$
05-18-2010, 12:20 PM
Post: #2
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
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