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    1. The problem statement, all variables and given/known data

     r_{A} (n) = number of solutions of  { \vec{x} \in Z^{m} ; A[\vec{x}] =n}

    where A[x]= x^t A x , is the associated quadratic from to the matrix A,

    where here A is positive definite, of rank m and even.

    (and I think symmetric?)

    I am solving for the r_{A}(1) for the two quadratic forms:

    Q(x,y,uv)= 2(x^2+y^2+u^2+v^2)+2xu+xv+yu-2yv

    R(x,y,uv)=x^2+4(y^2+u^2+v^2)+xu+  4yu+3yv+7uv

    2. Relevant equations

    see above

    3. The attempt at a solution

    diagonalized these read:

    Q=2(x+u/2+v/4)^2+2(y-v/2+u/4)^2+11u^2/8 + 11v^2/8

    R=(x+1/2u)^2+4(y+1/2u+3/8v)^2+11/4(u+v)^2+11/16v^2

    Solving Q=1 with all x,y,u,v integer, it is clear that u,v=0 is needed, and then x,y=\pm 1 gives r_{Q}(1)=4.

    Now looking at  r_{R}(1) by the same reasoning as above I would have said that we require v=0 , and then I' m not sure what to do.

    However the solution is:

    Must have u+v=1 & |v|=0,\pm 1 , this gives  \pm(1,0,0,0), \pm(1,0,-1,1) , \pm(0,0,-1,1)

    (the symbol that I interpreted as '&' in the solutions is a bit smudged, so looking at the solutions I'm not sure that this is supposed to be a 'or'? )

    Either way, I'm really confused, unsure where these conditions come from, how to think about this in a logical way...

    Many thanks for your help in advance.
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    (Original post by xfootiecrazeesarax)
    Many thanks for your help in advance.
    Usual caveat - not an expert.

    Can't say I follow your reasoning for Q, since u,v = 0 reduces your diagonalised form to Q=1=2x^2+2y^2 which has no integer solutions. Unless I'm being dense.

    Regarding R.
    Since R=1, and is the sum of squares, then each term must be <=1.

    So, (11/4) (u+v)^2 \leq 1

    For u,v integer this implies u+v=0, not 1 as given in the solutions.

    We also need (11/16)v^2 <=1, hence v must be one of -1,0,1.
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    (Original post by ghostwalker)
    Usual caveat - not an expert.

    Can't say I follow your reasoning for Q, since u,v = 0 reduces your diagonalised form to Q=1=2x^2+2y^2 which has no integer solutions. Unless I'm being dense.

    Regarding R.
    Since R=1, and is the sum of squares, then each term must be <=1.

    So, (11/4) (u+v)^2 \leq 1

    For u,v integer this implies u+v=0, not 1 as given in the solutions.

    We also need (11/16)v^2 <=1, hence v must be one of -1,0,1.
    thank you ! all makes alot more sense now
 
 
 
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