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  1. #1
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    help

    can anyone help me with these 2 super hard questions...

    1. If n is positive integer what is the number of solutions (x,y) (with x and y postive integers) to the equation
    1/x +1/y =1/n

    2. let p be an odd prime, and i be an integer, 0<= i <= p-2. show that 1^i +2^i +3^i +.....+ (p-1)^i = 0 (mod p)
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by felixmcgrady View Post
    can anyone help me with these 2 super hard questions...

    1. If n is positive integer what is the number of solutions (x,y) (with x and y positive integers) to the equation
    1/x +1/y =1/n

    2. let p be an odd prime, and i be an integer, 0<= i <= p-2. show that 1^i +2^i +3^i +.....+ (p-1)^i = 0 (mod p)
    1) is not too hard. If you multiply out the fractions, then the equation \tfrac1x+\tfrac1y = \tfrac1n can be written (x-n)(y-n)=n^2. So the number of solutions is the same as the number of ways of factorising n^2 as a product of two factors (which I'll leave you to think about).

    2) seems to be harder (unless I'm missing a simple solution). First, the question as stated is wrong, since the result is clearly false when i=0. So we want to prove it for 1\leqslant i \leqslant p-2.

    The only way I can see to do this is to quote the result that the multiplicative group \mathbb{Z}_p^* of nonzero elements of \mathbb{Z}_p is cyclic. Let r be a generator of this group. Then \sum_{k=1}^{p-1}k^i = \sum_{j=1}^{p-1}r^{ji}, because the elements r,r^2,\ldots,r^{p-1} are just 1,2,...,p-1 in some order. But if 1\leqslant i \leqslant p-2 then the map consisting of multiplication by i is a bijection of \mathbb{Z}_p^*. So the terms r^i,r^{2i},\ldots,r^{(p-1)i} are also just 1,2,...,p-1 in some order. That reduces the problem to showing that \sum_{k=1}^{p-1}k\equiv0\!\!\!\pmod p. But that is obviously true because \sum_{k=1}^{p-1}k = \tfrac12p(p-1).
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  3. #3
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    Quote Originally Posted by Opalg View Post
    (unless I'm missing a simple solution).
    That is the standard (simple) way of proving it.
    Except I just write out 1^i + ... + (p-1)^i \equiv r^i + r^{2i} + ... + r^{(p-1)i}
    And now apply geometric series.
    We from there that 1^i + ... + (p-1)^i \equiv 0 if  i \not | (p-1).
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