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Apr 6, 2010 at 22:56 comment added Bjorn Poonen In fact, what I was trying to hint at by "small integers" was your solution with k=4 (for p at least 5); i.e., 2^2 + 1^2 + 1^2 + ... + 1^2 + 0^2 + 0^2 = p. (No need to count quadratic residues!)
Apr 6, 2010 at 21:57 vote accept Kerry
Apr 6, 2010 at 23:36
Apr 6, 2010 at 21:46 history answered Michael Lugo CC BY-SA 2.5