Given any integers a, b, c and any prime p not a divisor of ab, prove that $\displaystyle ax^2+by^2\equiv c\ (mod\ p)$ is solvable.

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- Aug 22nd 2011, 01:23 PMalexmahoneProve that this equation is solvable
Given any integers a, b, c and any prime p not a divisor of ab, prove that $\displaystyle ax^2+by^2\equiv c\ (mod\ p)$ is solvable.

- Aug 24th 2011, 01:04 AMOpalgRe: Prove that this equation 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 $\displaystyle c\in\mathbb{Z}_p^*$ (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 $\displaystyle \mathbb{Z}_p^*$ belongs to each of the three sets R+R, R+N and N+N. But R is a subgroup of $\displaystyle \mathbb{Z}_p^*$. Its cosets are R and N. The sets$\displaystyle \{ax^2:x\in\mathbb{Z}_p^*\}$ and $\displaystyle \{by^2:y\in\mathbb{Z}_p^*\}$ are also cosets of R. Thus the set $\displaystyle \{ax^2 +by^2:x,y\in\mathbb{Z}_p^*\}$ is equal to one of the sets R+R, R+N or N+N, and therefore contains c.