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Math Help - Divisibility and perfect power

  1. #1
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    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)
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    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.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    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!
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    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
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by elim View Post
    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!
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  6. #6
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    Quote Originally Posted by elim View Post
    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?
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by mathman88 View Post
    I'm stuck with cases 1,2. Could you elaborate a bit?
    I'm a little hazy on the details too. Care to explain?
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  8. #8
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    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 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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