# Thread: Divisibility and perfect power

1. ## Divisibility and perfect power

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

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 = $\displaystyle 1^n$
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

5. 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 was making this much more complicated!

6. 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?

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 $\displaystyle a = b, \; k=\displaystyle{\frac{a^n+b^n}{(ab)^{n-1}+1}} \in \mathbb{N}^+ \Rightarrow a=b=k=1=1^n$
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 $\displaystyle \displaystyle{\frac{a^n+b^n}{(ab)^{n-1}+1}=\frac{a^n(1+a^{2n(n-1)})}{(a^{2n})^{n-1}+1}} = a^n$