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 (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 belongs to each of the three sets R+R, R+N and N+N. But R is a subgroup of . Its cosets are R and N. The sets and are also cosets of R. Thus the set is equal to one of the sets R+R, R+N or N+N, and therefore contains c.