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    Find all non negative integers n such that 5^n-4^n is divisible by 61

    So after doing some calculations i guessed that any multiple of 3 would stastify it. So i need to show that firstly n=3k satisfies it and that n=3k+1 and n=3k+2 doesn't

    so by using the factor theorem

    (5^3-4^3) is a factor of 5^(3k)-4^(3k) hence 61 is a factor in the case n=3k

    if n=3k+1 then


    5^(3k)*(4+1)-5^(3k)*4=4(5^(3k)-4^(3k))+5^(3k) and since 5^3k always ends in a 5 61 cannot divide it.

    in the case n=3k+2

    5^(3k)*(4^2+3^2)-4^(3k)*4^2=4^2(5^(3k)-4^(3k)-3^2*5^(3k) if 61 is to divide this then it must be possible to decompose 61 into a product of 5's or 3's or both. Since this is not possible 5^n-4^n is not divisible by 61 in the case n=3k+2

    is this correct?
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    sorry if im being dumb, but where did you show that 5^3-4^3|5^{3k}-4^{3k} or is this something so trivial I'm being a twit?

    the rest of it seems fine.

    it would be a lot easier to read if you latexed it mind you...

    I think I would use induction on this.
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    (Original post by Totally Tom)
    sorry if im being dumb, but where did you show that 5^3-4^3|5^{3k}-4^{3k} or is this something so trivial I'm being a twit?

    the rest of it seems fine.

    it would be a lot easier to read if you latexed it mind you...

    I think I would use induction on this.
    (x^a - y^a)|(x^ab - y^ab) for all integral x, y, a, b .

    Proof:

    (x^ab - y^ab) = (x^a - y^a)[x^a(b - 1) + x^a(b - 2)*y^b + ... + x^b*y^a(b - 2) + y^a(b - 1)].

    Hope this clears it up.

    ~~Simba

    Edit: Sorry, slight typo!
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    yeah that's what i did, sorry for not making it clearer
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    (Original post by Simba)
    (x^a - y^a)|(x^ab - y^ab) for all integral x, y, a, b .

    Proof:

    (x^ab - y^ab) = (x^a - y^a)[x^a(b - 1) + x^a(b - 2)*y^b + ... + x^b*y^a(b - 2) + y^a(b - 1)].

    Hope this clears it up.

    ~~Simba

    Edit: Sorry, slight typo!
    How did you get that?

    sorry lol, couldn't follow it

    Edit: I stopped being a lazy **** and got it . Is that some sort of standard result, or did you calculate that yourself!
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    well, you should know that x^n-1=(x-1)x^n+x^n-1...+1) it's sort of an extension of it i guess
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    Yes, it arises from the factor theorem.

    If p(x) is a polynomial, and p(a)=0 then (x-a) is a factor.

    so in p(a)=a^n-b^n (a-b) is a factor
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    (Original post by n1r4v)
    Is that some sort of standard result, or did you calculate that yourself!
    It is a standard olympiad result .
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    (Original post by Simba)
    It is a standard olympiad result .
    Ah ok thanks very much

    What is this olympiad thing anyway? Is it the "maths challenge" thing?
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    (Original post by n1r4v)
    Ah ok thanks very much

    What is this olympiad thing anyway? Is it the "maths challenge" thing?
    well you do the maths challenge, then you go onto the first round of the BMO, british maths olyimpiad then there is BMO2 and then the IMO international...

    they are all progressively harder, i can't do anything in BMO2.
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    (Original post by n1r4v)
    Ah ok thanks very much

    What is this olympiad thing anyway? Is it the "maths challenge" thing?
    See the following link for the (relatively) easy BMO1 papers and tougher BMO2 papers:

    http://bmoc.maths.org.uk/home/bmo.shtml

    After this there are more tests before the IMO, but they are not publically available.
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    Cheers for the info guys
 
 
 
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