Divisibility and perfect power

• May 14th 2010, 02:48 PM
elim
Divisibility and perfect power
If $\displaystyle a,b,n \in \mathbb{N}^+, \quad n \ge 2$
Prove that $\displaystyle (k = \frac{a^n+b^n}{(ab)^{n-1}+1} \in \mathbb{N}) \Rightarrow \exists c \in \mathbb{N} \; (k = c^n)$
• May 17th 2010, 06:05 PM
chiph588@
This a stupendous problem! (Clapping) May I ask where you got this from?

I would love to take a crack at this right now but I'm so busy at the moment, but what I can offer you is this. Hopefully things will clear up for me in the coming week and I'll have a go at it.
• May 17th 2010, 07:39 PM
Bruno J.
I'm going to bed now, but I'll just remark that, in general, the denominator is much bigger than the numerator. There may in fact be very few cases to check!
• May 18th 2010, 05:36 PM
elim
The basics
When a = b, one can easily see that a = b = k = 1 = $\displaystyle 1^n$
Assume that $\displaystyle 1 \le a < b$ and $\displaystyle k \in \mathbb{N}^+$ satisties $\displaystyle a^n - k = b^{n-1} (k a^{n-1} -b)$
Consider 3 cases:
(1) $\displaystyle k > a^n$ (2) $\displaystyle k < a^n$ and (3) $\displaystyle k = a^n$
Try to prove that (1) leads to contradiction (2) forces $\displaystyle k=1$ and (3) gives non-trivial (a,b,k)'s
• May 18th 2010, 06:41 PM
chiph588@
Quote:

Originally Posted by elim
When a = b, one can easily see that a = b = k = 1 = $\displaystyle 1^n$
Assume that $\displaystyle 1 \le a < b$ and $\displaystyle k \in \mathbb{N}^+$ satisties $\displaystyle a^n - k = b^{n-1} (k a^{n-1} -b)$
Consider 3 cases:
(1) $\displaystyle k > a^n$ (2) $\displaystyle k < a^n$ and (3) $\displaystyle k = a^n$
Try to prove that (1) leads to contradiction (2) forces $\displaystyle k=1$ and (3) gives non-trivial (a,b,k)'s

(Clapping) I was making this much more complicated!
• May 18th 2010, 08:44 PM
mathman88
Quote:

Originally Posted by elim
When a = b, one can easily see that a = b = k = 1 = $\displaystyle 1^n$
Assume that $\displaystyle 1 \le a < b$ and $\displaystyle k \in \mathbb{N}^+$ satisties $\displaystyle a^n - k = b^{n-1} (k a^{n-1} -b)$
Consider 3 cases:
(1) $\displaystyle k > a^n$ (2) $\displaystyle k < a^n$ and (3) $\displaystyle k = a^n$
Try to prove that (1) leads to contradiction (2) forces $\displaystyle k=1$ and (3) gives non-trivial (a,b,k)'s

I'm stuck with cases 1,2. Could you elaborate a bit?
• May 29th 2010, 04:46 PM
chiph588@
Quote:

Originally Posted by mathman88
I'm stuck with cases 1,2. Could you elaborate a bit?

I'm a little hazy on the details too. Care to explain? (Happy)
• May 30th 2010, 02:53 PM
elim
ref

Proof We have $\displaystyle a = b, \; k=\displaystyle{\frac{a^n+b^n}{(ab)^{n-1}+1}} \in \mathbb{N}^+ \Rightarrow a=b=k=1=1^n$
Now assume that $\displaystyle a<b,\;n>1,\; a^n-k=(ka^{n-1}-b)b^{n-1}$
(1) If $\displaystyle k > a^n$ then $\displaystyle k > k-a^n = b^{n-1} (b-ka^{n-1}) \ge b^{n-1}$
$\displaystyle \quad \;$ thus $\displaystyle b > ka^{n-1} \ge k > b^{n-1}$. It's impossible.
(2) If $\displaystyle k < a^n$, then $\displaystyle ka^{n-1}-b = \displaystyle{\frac{a^n-k}{b^{n-1}}} < a$, $\displaystyle ka^{n-1} < a+b$, $\displaystyle (k-1)a^{n-1} < a+b - a^{n-1} \le b$
$\displaystyle \quad$ If $\displaystyle k > 1$, then $\displaystyle a^{n-1} \le (k-1)a^{n-1} < b$, $\displaystyle a^n-k = (ka^{n-1}-b)b^{n-1} \ge b^{n-1} > (a^{n-1})^{n-1} \ge a^n$,
$\displaystyle \quad \;$ that's wrong and so $\displaystyle k=1=1^n$
(3) If $\displaystyle k = a^n$ then $\displaystyle b=ka^{n-1}=a^{2n-1} \; b^n = a^{2n^2-n} = a^{2n(n-1) + n}$ hence $\displaystyle \displaystyle{\frac{a^n+b^n}{(ab)^{n-1}+1}=\frac{a^n(1+a^{2n(n-1)})}{(a^{2n})^{n-1}+1}} = a^n$