## Quadratic Residues – Part III

Ok, here’s the third installation. Getting a little tired of repeatedly saying “a is/isn’t a square mod p“, we introduce a new notation.

Definition. Let p be an odd prime and a be an integer coprime to p. The Legendre symbol $(\frac a p)$ is given by: $(\frac a p) = \begin{cases} +1, \quad &\mbox{if } a \mbox{ is a square mod p},\\ -1,\quad &\mbox{if } a \mbox{ is not a square mod p}.\end{cases}$

For completeness, one also defines $(\frac a p) = 0$ if a is a multiple of p, but we won’t need this for now. From our previous post, we know the following:

1. $(\frac a p) \equiv a^{(p-1)/2} \pmod p$;
2. hence from property 1, $(\frac a p) (\frac b p) = ( \frac {ab} p )$;
3. for special values of a, we have $(\frac {-1} p) = (-1)^{(p-1)/2}$ and $(\frac 2 p) = (-1)^{(p^2-1)/8}$.

From the multiplicative property (2), it thus suffices to consider $(\frac q p)$ for odd prime q ≠ p. This is where we’ll pull the next rabbit out of the hat (i.e. quote a result without proof).

Quadratic Reciprocity Theorem. If p, q are distinct odd primes, then $(\frac p q) (\frac q p) = (-1)^{(p-1)(q-1)/4}$.

In other words, $(\frac p q)$ and $(\frac q p)$ are identical unless both p and q are congruent to 3 mod 4.

Example 1. Determine whether 31 is a square modulo 73.

Solution. We will repeatedly use the quadratic reciprocity rule:

• $(\frac{31}{73}) (\frac{73}{31}) = +1$ since 73 is 1 mod 4, so $(\frac{31}{73}) = (\frac{73}{31}) = (\frac{11}{31})$;
• $(\frac{11}{31}) (\frac{31}{11}) = -1$ since 31 ≡ 11 ≡ 3 (mod 4), so $(\frac{11}{31}) = -(\frac{31}{11}) = -(\frac {9}{11}) = -1$ since 9 is clearly a square mod 11.

Thus, 31 is not a square modulo 73. ♦

Example 2. Classify all odd primes p ≠ 3 such that 3 is a square modulo p.

Solution. Use the quadratic reciprocity: $(\frac 3 p) (\frac p 3) = (-1)^{(p-1)/2}$. Since the RHS depends on p modulo 4, we need to consider p modulo 12. Thus, 3 is a square modulo p iff $(\frac p 3) = (-1)^{(p-1)/2}$.

• If $(\frac p 3) = (-1)^{(p-1)/2} = +1$, then $p \equiv 1 \pmod 3$ and $p \equiv 1 \pmod 4$, i.e. $p \equiv 1 \pmod {12}$.
• If $(\frac p 3) = (-1)^{(p-1)/2} = -1$, then $p \equiv 2 \pmod 3$ and $p \equiv 3 \pmod 4$, i.e. $p \equiv 11 \pmod {12}$.

Thus 3 is a square mod p iff p ≡ ±1 (mod 12). ♦

Example 3. Classify all odd primes p ≠ 3 such that -3 is a square modulo p.

Solution. From example 1, we have $(\frac 3 p) = (\frac p 3) (-1)^{(p-1)/2}$. Thus, $(\frac {-3} p) = (\frac {-1} p) (\frac 3 p) = (-1)^{(p-1)/2}(\frac 3 p)=(\frac p 3)$.

And -3 is a square mod p iff p ≡ 1 (mod 3). ♦

Coming up next: applying quadratic residues in conjunction with order of an element modulo n to solve IMO-type problems. The gist of the idea is this: suppose p is a prime and a be not divisible by p; let the (multiplicative) order of a modulo p be denoted d. We already know that d | (p-1). It turns out that a is a square modulo p if and only if $\frac {p-1} d$ is even.

To see why, just let g be a primitive root modulo p. And write $a \equiv g^r \pmod p$. Now d is the smallest positive integer for which $a^d \equiv g^{rd} \equiv 1 \pmod p$. Thus d is the smallest positive integer such that rd is a multiple of (p-1). A moment of thought will tell you that thus d = (p-1)/gcd(r, p-1) (prove it! just write g = gcd(r, p-1) and r = gu, (p-1) = gv where (u, v) = 1 … ).

Thus, our desired result follows from the fact that a is a square modulo p if and only if r is even. If you’re not sure why the fact is true, refer to part II of the notes. If you’re still confused, fret not, we will include some concrete examples in the next installation.

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