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Thread: Set of residues

  1. #1
    Super Member Showcase_22's Avatar
    Sep 2006
    The raggedy edge.

    Set of residues

    Is there a complete set of residues modulo p consisting of perfect squares:

    i). When p=7.

    ii). for any prime p.
    I can probably work out part ii). if I can do part i).

    I think part i) is true but I can't figure out the set. My set so far is \{0,36,9\} for 0,1 and 2 respectively and \{25 \} for 4.

    I can't find any perfect squares for 3,5 and 6!
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  2. #2
    Senior Member
    Apr 2009
    Atlanta, GA

    There is no such set ever

    Consider the sequence of squares 0,1,4,9,16,25,36,49,64,81,100,... in modulo:

    \bmod 7: 0,1,4,2,2,4,1,0,1,4,2,2,4,1,0,1,... (notice a pattern?)

    Notice that (n+7)^2-n^2=14n+7 \equiv 0 (\bmod 7), so the set repeats in blocks of 7.

    Also, (7k+r)^2-(7k-r)^2=28kr \equiv 0 (\bmod 7) so the set "reflects" as shown in this example. More clearly, for any k, (7k)^2 \equiv 0 (\bmod 7), therefore (7k+r)^2 \equiv (7k-r)^2 (mod 7) for all r.

    Both of these observations can be generalized. For any prime p,

    (n+p)^2-n^2 \equiv 0 (\bmod p)

    (pk+r)^2-(pk-r)^2 \equiv 0 (\bmod p)

    For all appropriate n,k,r

    Therefore, the size of the set of residues of n^2 for any prime p is at most \lceil \frac{p}{2} \rceil

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