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Math Help - Quadratic Residues

  1. #1
    Super Member PaulRS's Avatar
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    Quadratic Residues

    Let p be a prime such that p>(n^2+n+k)^2+k. ( <br />
n,k \in \mathbb{Z}^ +  <br />
)

    Consider the sequence: <br />
n^2 ,n^2  + 1,...,\left( {n^2  + n + k} \right)^2  + k<br />
.

    Prove that there's a pair (m,m+k) of integers from our sequence, such that <br />
\left( {\tfrac{m}<br />
{p}} \right)_L  = \left( {\tfrac{{m + k}}<br />
{p}} \right)_L  = 1<br />

    This is a lovely problem, have fun!.

    PS: By the way, I'd posted 2 problems like in January, NCA gave a solution for 1, but the other is waiting! (haha, and it is a nice problem too)
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  2. #2
    MHF Contributor

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    Quote Originally Posted by PaulRS View Post
    Let p be a prime such that p>(n^2+n+k)^2+k. ( <br />
n,k \in \mathbb{Z}^ + <br />
)

    Consider the sequence: <br />
n^2 ,n^2 + 1,...,\left( {n^2 + n + k} \right)^2 + k<br />
.

    Prove that there's a pair (m,m+k) of integers from our sequence, such that <br />
\left( {\tfrac{m}<br />
{p}} \right)_L = \left( {\tfrac{{m + k}}<br />
{p}} \right)_L = 1<br />

    This is a lovely problem, have fun!.

    PS: By the way, I'd posted 2 problems like in January, NCA gave a solution for 1, but the other is waiting! (haha, and it is a nice problem too)
    if \left(\frac{n^2+k}{p} \right)=1, choose m=n^2 and if \left(\frac{(n+1)^2+k}{p} \right)=1, then choose m=(n+1)^2. so we may assume that \left(\frac{n^2+k}{p} \right)=\left(\frac{(n+1)^2+k}{p} \right)=-1.

    now we have this cute identity: (n^2+k)((n+1)^2+k)=(n^2+n+k)^2+k. therefore: \left(\frac{(n^2+n+k)^2+k}{p} \right)=1 and we choose m=(n^2+n+k)^2. \ \Box



    Remark: to put the above result in simple words: for any positive integers n,k and any prime number p > (n^2+n+k)^2+k, at least one of the numbers

    n^2+k, \ (n+1)^2+k, \ (n^2+n+k)^2+k is a quadratic residue modulo p.
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