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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