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Find all integers with b=a^n such that a^2 + b^2 is divisible by ab+1

  1. Jul 6, 2008 #1
    Let [itex]a[/itex] be a positive integer. Find all positive integers [itex]n[/itex] such that [itex]b = a^n[/itex] satisfies the condition that [itex]a^2 + b^2[/itex] is divisible by [itex]ab + 1[/itex].

    Obviously if [itex]a=1[/itex] then all [itex]n[/itex] work. Otherwise, we have [itex]a^2 + b^2 = a^2 (1+a^{2(n-1)})[/itex]. Also, [itex]a^2[/itex] and [itex]a^{n+1} + 1[/itex] are relatively prime, so we need to find all [itex]n[/itex] such that [itex]a^{n+1} + 1[/itex] divides [itex]1+a^{2(n-1)}[/itex]. Clearly [itex]n=3[/itex] works, but now I'm stuck. What do I do now?
     
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  3. Jul 6, 2008 #2

    morphism

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    Re: Divisibility

    What can you say about the polynomial [itex]x^{2k} + 1[/itex]?
     
  4. Jul 6, 2008 #3
    Re: Divisibility

    I don't know. It has no real zeros? That doesn't mean it can't be factored. For example 2^2+1|2^6+1.
     
  5. Jul 6, 2008 #4

    morphism

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    Re: Divisibility

    Right, but there's something special we can say about its factors. Here's another hint: What can you say about [itex]x^{2^m k} + 1[/itex] when k is odd?
     
  6. Jul 6, 2008 #5
    Re: Divisibility

    Let [itex]k=2j+1[/itex]. Then [itex]x^{2^m (2j+1)} + 1 = (x^{2^m}+1)(x^{2^m (2j)} - x^{2^m (2j - 1)} + ... + 1)[/itex]. So [itex]x^{2^m}+1 | x^{2^m k} + 1[/itex]. I'm still stuck :smile:.
     
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