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I'm trying to understand what is established in Proofs from the Book (from Eigner and Ziegler) concerning the representation of numbers as a sum of two squares.

Consider $\mathbb{Z}_p$, $p$ an odd prime. At first, the authors affirm that is always possible to partition $\{1,2, \dots, p-1\}$ in quadruples $\{x, -x, \overline{x}, - \overline{x}\}$, where $-x$ is the additive inverse of $x$ and $\overline{x}$ is the multiplicative inverse of $x$. If two of these numbers are equal, the class reduces to a pair of elements. I proved the last affirmation.

My doubt is: the authors affirm that, if $p-1=4m+2$ we will have just one pair and the other elements will be quadruples and, if $p-1=4m$, there will be 2 pairs. According to the number of these pairs, they establish the number of solutions of $x^2 \equiv 1 \mod p$.

All my examples indicates that it is true (of course it is because the book is a classic!), but I can't understand it.

Jordan Green
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Cleto Pereira
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1 Answers1

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If $p-1 = 4m+2$ for some $m$, then the number of pairs must be odd. (Otherwise, after taking the union of all quadruples and pairs, we’d get a total number of elements that was divisible by $4$.) Similarly, if $p−1=4m$, then the number of pairs must be even. Thus, to prove the authors’ claim, it suffices to prove that there are at most two pairs, regardless of the value of $p-1$ modulo $4$.

The key to proving that there are at most two pairs is the fact that if $p$ is prime, then a degree $n$ polynomial can have at most $n$ roots modulo $p$. This can be proved via induction. (See, e.g. If $f$ is a polynomial of degree $n$, then $f(x) \equiv 0\pmod p$ has at most $n$ solutions.). Here, the relevant polynomials are $f(x) = x^2 -1$ and $g(x) = x^2+1$, because a quadruple $\{x,-x, \overline{x}, - \overline{x} \}$ reduces to a pair if and only if $x$ and $-x$ are both roots of $f$ or they are both roots of $g$. You probably proved something like this in the course of proving that if two elements of $\{x,-x, \overline{x}, - \overline{x} \}$ are equal, then it reduces to a pair; however, I’ve appended a proof for the sake of completeness.*

Since $f$ and $g$ each have at most $2$ roots mod $p$, at most one quadruple can be reduced to a pair by way of solving each of $f$ and $g$. Thus, at most two quadruples can be reduced to pairs.

We note that $f$ always has two solutions mod $p$, namely $1$ and $p-1$. Thus, the one pair we get when $p = 4m+1$ must result from roots of $f$, not roots of $g$. (The roots of $f$ and $g$ never coincide assuming that $p$ is odd.)

*Appendix: Proof that $\{x,-x, \overline{x}, - \overline{x} \}$ reduces to a pair if and only if $x$ and $-x$ solve $f$ or $x$ and $-x$ solve $g$.

By setting pairs from $\{x,-x, \overline{x}, - \overline{x} \}$ equal, one obtains six equations: $$ x = -x \tag{1}; $$

$$ x = \overline{x}; \tag{2} $$

$$ x = - \overline{x}; \tag{3} $$

$$ -x = \overline{x}; \tag{4} $$

$$ -x = - \overline{x}; \tag{5} $$

$$ \overline{x} = -\overline{x}. \tag{6} $$

However, (1) and (6) are impossible for an odd $p$, while (2) is equivalent to (5) and (3) is equivalent to (4). Thus, the quadruple reduces to a pair if and only if (2) is satisfied or (3) is satisfied. In turn, (2) is satisfied if and only if $x$ and $-x$ are both roots of $f$, whereas (3) is satisfied if and only if $x$ and $-x$ are roots of $g$.

Jordan Green
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