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Math Help - Prove that this equation is solvable

  1. #1
    MHF Contributor alexmahone's Avatar
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    Prove that this equation is solvable

    Given any integers a, b, c and any prime p not a divisor of ab, prove that ax^2+by^2\equiv c\ (mod\ p) is solvable.
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    MHF Contributor
    Opalg's Avatar
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    Re: Prove that this equation is solvable

    Quote Originally Posted by alexmahone View Post
    Given any integers a, b, c and any prime p not a divisor of ab, prove that ax^2+by^2\equiv c\ (mod\ p) is solvable.
    I can only prove this by quoting a theorem from a recent research paper. However, the theorem is not that hard. Essentially the same result seems to have been proved a long time ago, in this 1893 article by J C Fields (after whom the Fields Medal is named).

    If c is a multiple of p then we can take x = y = 0. So we can assume that c\in\mathbb{Z}_p^* (the multiplicative group of nonzero residues mod p). Also, the result is easy to prove in the case p=2. So assume that p is an odd prime.

    Let R be the set of quadratic residues mod p, and let N be the set of quadratic non-residues. According to Theorem 2.1 in this paper (pdf file), every element of \mathbb{Z}_p^* belongs to each of the three sets R+R, R+N and N+N. But R is a subgroup of \mathbb{Z}_p^*. Its cosets are R and N. The sets  \{ax^2:x\in\mathbb{Z}_p^*\} and \{by^2:y\in\mathbb{Z}_p^*\} are also cosets of R. Thus the set \{ax^2 +by^2:x,y\in\mathbb{Z}_p^*\} is equal to one of the sets R+R, R+N or N+N, and therefore contains c.
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