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    Does (n + a) | (bn + (a-c)), where a, b and c are known, a,b,c, \in \mathbb{N}, and n \not = 0, a\not = 0, a\not = c, ever have any solutions when gcd(b,a-c) = 1?

    For example, could I infer that (n+2008)|(2n + 2007) can only be true for n=c=1 or not? (Or is it even true?)
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    Well, I would first deduce that n + 2008 divides n - 1. But n + 2008 > n - 1, so we conclude that n \le 1, since there are no multiples of n + 2008 between 0 and n - 1. But n \in \mathbb{N}, so n = 1.
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    Thank you, that completes my rather shoddy proof, but is it true in general?
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    Somehow I've made it through infant school, primary school, secondary school and up to my A2s without knowing how to do division on paper... so looking at that makes me brick myself
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    (Original post by rupertj)
    Does (n + a) | (bn + (a-c)), where a, b and c are known, a,b,c, \in \mathbb{N}, and n \not = 0, a\not = 0, a\not = c, ever have any solutions when gcd(b,a-c)\not = 1?
    n + 8 | 2n + 4 (a = 8, b = 2, c = 4, so gcd(b, a-c) = 2) has a solution n = 4.
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    (Original post by generalebriety)
    n + 8 | 2n + 4 (a = 8, b = 2, c = 4, so gcd(b, a-c) = 2) has a solution n = 4.
    Apologies, I meant gcd(b,a-c)=1, rather than 'is not equal to'.

    I've edited it now.
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    (Original post by rupertj)
    Apologies, I meant gcd(b,a-c)=1, rather than 'is not equal to'.

    I've edited it now.
    Ok. n+7|3n+2 (a=7, b=3, c=5, so gcd(b, a-c) = 1) has solution n=12.
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    (Original post by generalebriety)
    Ok. n+7|3n+2 (a=7, b=3, c=5, so gcd(b, a-c) = 1) has solution n=12.
    Okay, thanks. I'm so stupid.
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    (Original post by rupertj)
    Okay, thanks. I'm so stupid.
    In fact, here's something a bit stronger. Suppose we want to solve n+a | bn + (a-c). Then one solution obviously occurs if (n+a) divides (bn + (a-c)) precisely (b-1) times. That is:

    (b-1)(n+a) = bn + a - c
    i.e.
    (b-1)n + a(b-1) = bn + a - c
    and so
    n = a(b-2) + c

    which is always a positive integer if b >= 2.
 
 
 

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