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How can I prove that a sum of two integer squares, namely $ x^2 + y^2 $ (ranging from $ x = 0 \to p, \; y = 0 \to p $) is a complete system of residues (CSR) modulo $ p $ (prime)? Or, how can I prove that $ x^2 + y^2 \equiv \left \{ 0, 1, 2, \cdots, p-1 \right \} \pmod p $, for some integers $ x,y $ and all prime values of $ p $?

I've failed to prove $ \left (0, 1, 4, 9, 16, \cdots \right ) $ is a CSR modulo $ p $ , and I feel that this is an important step to solve my question.

I've also tried showing that Lagrange's four-square theorem solves the question, but it obviously gets stuck when $ x^2 + y^2 \equiv z \pmod p $, and $ z $ cannot be expressed as a sum of squares.

Dela Corte
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    It's unclear what you mean by a sum of two integer squares being a complete system. A sum is a single number. You need to specify the range of $x$ and $y$ or it makes no sense to ask if it's a complete system of residues.

    Consider the prime $p=5$. The square sequence you mentioned is here $0, 1, 4$. This is not a complete system of residues modulo $p$

     – Zavosh Sep 30 '14 at 11:20
  • @Prometheus Sorry for being unclear. The square sentence for $ p = 5 $ should be read as $ (0+0, 1+0, 2^2+0, 0+1, 1+1, 1+2^2, 0+2^2, 1+2^2, 2^2+2^2, \cdots) $, and applying modulo $ 5 $ the sequence becomes $ (0, 1, 2, 3, 4 ) $, and thus it is a complete system of residues. – Dela Corte Sep 30 '14 at 11:50

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You should be able to show that precisely half the non-zero positive integers less than $p$ are squares modulo $p$. (This is because the polynomial $x^2 - c$ has at most $2$ roots in the field $\mathbb{Z}/p\mathbb{Z}$).

Let $n$ be the smallest integer such that $n$ is not a square modulo $p$. Then $na^2$ (modulo $p$) is also not a square modulo $p$ for $a \neq 0$, since if $na^2 = b^2$ then $n = (ba^{-1})^2$. But the minimality of $n$ tells us that $n-1 = b^2$ for some $a \neq 0$. So $na^2 = a^2 + (ba)^2$. But by counting, every non-square must be of the form $na^2$ for some $a$, and so of the form $x^2 + y^2$. And the squares are obviously of this form, so we're done.

Christopher
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