# Thread: Divisibility and perfect power

1. ## Divisibility and perfect power

If $a,b,n \in \mathbb{N}^+, \quad n \ge 2$
Prove that $(k = \frac{a^n+b^n}{(ab)^{n-1}+1} \in \mathbb{N}) \Rightarrow \exists c \in \mathbb{N} \; (k = c^n)$

2. This a stupendous problem! 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.

3. 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!

4. ## The basics

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

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

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

7. 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?

8. ref

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 $a1,\; a^n-k=(ka^{n-1}-b)b^{n-1}$
(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 $\displaystyle{\frac{a^n+b^n}{(ab)^{n-1}+1}=\frac{a^n(1+a^{2n(n-1)})}{(a^{2n})^{n-1}+1}} = a^n$