Legendre符号求和等式∑_{y=0}^{p-1 } (rac{y² - b⁻¹}{p}) = -1的详细证明请求

Alright, let's break this down step by step. I'll assume ( p ) is an odd prime (the standard context for this identity—we can verify the ( p=2 ) case separately later) and ( b ) is invertible modulo ( p ) (so ( b^{-1} ) exists, meaning ( b \not\equiv 0 \pmod{p} )). First, let's simplify notation by setting ( c = b^{-1} \pmod{p} ); our goal becomes proving:
$$
\sum_{y=0}^{p-1} \left( \frac{y^2 - c}{p} \right) = -1
$$

Step 1: Rewrite the sum using quadratic equation solution counts #

Let ( \chi(x) = \left( \frac{x}{p} \right) ) denote the Legendre symbol (our quadratic character modulo ( p )). For each ( x \in \mathbb{Z}/p\mathbb{Z} ), let ( N(x) ) be the number of solutions ( y ) to ( y^2 = x + c \pmod{p} ).

We can rephrase our original sum by grouping terms by the value of ( x = y^2 - c ):
$$
\sum_{y=0}^{p-1} \chi(y^2 - c) = \sum_{x=0}^{p-1} \chi(x) \cdot N(x)
$$
Each ( \chi(x) ) is multiplied by how many times ( x ) appears as ( y^2 - c ) (i.e., how many ( y ) map to that ( x )).

Step 2: Characterize ( N(x) ) with the Legendre symbol #

For any ( d = x + c \pmod{p} ), the quadratic equation ( y^2 = d ) has:

• Exactly 1 solution if ( d \equiv 0 \pmod{p} ) (only ( y=0 ))
• 2 solutions if ( d ) is a non-zero quadratic residue
• 0 solutions if ( d ) is a non-quadratic residue

This count can be written compactly using the Legendre symbol:
$$
N(x) = 1 + \chi(x + c)
$$

• When ( x + c = 0 ) (i.e., ( x = -c )), ( N(x) = 1 = 1 + \chi(0) = 1 + 0 ), which holds.
• When ( x + c \neq 0 ), ( \chi(x + c) = 1 ) (residue) gives ( N(x)=2 ), and ( \chi(x + c)=-1 ) (non-residue) gives ( N(x)=0 )—perfectly matching our earlier count.

Step 3: Expand and simplify the sum #

Substitute ( N(x) = 1 + \chi(x + c) ) into our sum:
$$
\sum_{x=0}^{p-1} \chi(x) \cdot N(x) = \sum_{x=0}^{p-1} \chi(x) + \sum_{x=0}^{p-1} \chi(x) \cdot \chi(x + c)
$$

Let's compute each part separately:

1. First sum: ( \sum_{x=0}^{p-1} \chi(x) = 0 )

   • The Legendre symbol is 0 for ( x=0 ), 1 for the ( (p-1)/2 ) quadratic residues, and -1 for the ( (p-1)/2 ) non-residues. The positive and negative terms cancel exactly, so the total is 0.

2. Second sum: ( \sum_{x=0}^{p-1} \chi(x) \cdot \chi(x + c) = \sum_{x=0}^{p-1} \chi(x(x + c)) )

   • We use the multiplicative property of the Legendre symbol: ( \chi(ab) = \chi(a)\chi(b) ).

Now simplify this second sum:

• For ( x=0 ): ( \chi(0 \cdot (0 + c)) = \chi(0) = 0 ), so this term contributes nothing.
• For ( x \neq 0 ): Factor out ( x^2 ) from the argument:
  $$
  \chi(x(x + c)) = \chi\left(x^2\left(1 + c x^{-1}\right)\right) = \chi(x^2) \cdot \chi\left(1 + c x^{-1}\right)
  $$
  Since ( x^2 ) is always a quadratic residue, ( \chi(x^2) = 1 ). Let ( z = x^{-1} \pmod{p} ): as ( x ) runs over non-zero elements, ( z ) also runs over all non-zero elements. The sum becomes:
  $$
  \sum_{z=1}^{p-1} \chi(1 + c z)
  $$

Let ( w = 1 + c z ). As ( z ) runs over non-zero elements, ( c z ) runs over all non-zero elements (since ( c \neq 0 )), so ( w ) runs over all elements of ( \mathbb{Z}/p\mathbb{Z} ) except ( w=1 ) (because ( c z \neq 0 \implies w = 1 + c z \neq 1 )). Therefore:
$$
\sum_{z=1}^{p-1} \chi(1 + c z) = \sum_{w=0}^{p-1} \chi(w) - \chi(1)
$$
We know ( \sum_{w=0}^{p-1} \chi(w) = 0 ), and ( \chi(1) = 1 ) (1 is a quadratic residue). So this sum equals ( 0 - 1 = -1 ).

Step 4: Combine the results #

Putting it all together:
$$
\sum_{y=0}^{p-1} \left( \frac{y^2 - c}{p} \right) = 0 + (-1) = -1
$$
Since ( c = b^{-1} ), this is exactly the identity we needed to prove.

Quick check for ( p=2 ) #

If ( p=2 ), ( b ) must be 1 (the only invertible element), so ( b^{-1}=1 ). The sum runs over ( y=0,1 ):

• ( y=0 ): ( \left( \frac{0 - 1}{2} \right) = \left( \frac{1}{2} \right) = 1 )
• ( y=1 ): ( \left( \frac{1 - 1}{2} \right) = \left( \frac{0}{2} \right) = 0 )
  Sum is 1, but note the original identity assumes ( p ) is an odd prime (the standard case for such Legendre symbol identities).

内容的提问来源于stack exchange，提问作者User432477438