Investigate the properties of a sum of complex roots of unity where the base is restricted to quadratic residues modulo a prime number | Step-by-Step Solution

Understanding the Gauss Sum for Quadratic Residues

What We're Solving:

We want to find the sum ∑ e^(2πik/p) where k runs over all quadratic residues modulo an odd prime p. This is asking us to add up specific complex roots of unity - but only those corresponding to numbers that are perfect squares mod p.

The Approach:

This problem combines two beautiful areas of mathematics! We're using:

• Complex analysis (roots of unity on the unit circle)
• Number theory (quadratic residues and the Legendre symbol)

The key insight is to use the Legendre symbol (k/p) to "filter" our sum, picking out only the quadratic residues. Then we'll use properties of character sums to evaluate it.

Step-by-Step Solution:

Step 1: Set up the filtering mechanism Let's use the Legendre symbol (k/p), which equals:

• +1 if k is a quadratic residue mod p
• -1 if k is a quadratic non-residue mod p
• 0 if k ≡ 0 (mod p)

Our sum becomes: S = ∑_{k=1}^{p-1} (1 + (k/p))/2 · e^(2πik/p)

The factor (1 + (k/p))/2 equals 1 when k is a quadratic residue and 0 otherwise - perfect!

Step 2: Expand and separate S = (1/2)[∑_{k=1}^{p-1} e^(2πik/p) + ∑_{k=1}^{p-1} (k/p)e^(2πik/p)]

Step 3: Evaluate the first sum The first sum is just all the p-th roots of unity except 1: ∑_{k=1}^{p-1} e^(2πik/p) = -1

This is because all p-th roots of unity sum to zero!

Step 4: Recognize the Gauss sum The second sum is the famous Gauss sum: G = ∑_{k=1}^{p-1} (k/p)e^(2πik/p)

Step 5: Use the known value of the Gauss sum From the theory of quadratic Gauss sums, we know:

• If p ≡ 1 (mod 4), then G = √p
• If p ≡ 3 (mod 4), then G = i√p

Step 6: Put it all together S = (1/2)[-1 + G] = (1/2)(-1 + (√p or i√p))

The Answer:

For an odd prime p, the sum equals:

• (√p - 1)/2 if p ≡ 1 (mod 4)
• (i√p - 1)/2 if p ≡ 3 (mod 4)

Notice how beautifully this connects geometry (complex numbers) with number theory (quadratic residues)!

Memory Tip:

Think of this as "half the Gauss sum, shifted by the missing root of unity." The quadratic residues are exactly half the non-zero elements mod p, so we get roughly half of what we'd expect from the full Gauss sum, adjusted for the geometric series of roots of unity.

Great job tackling this advanced topic - you're working with some of the most elegant mathematics ever discovered! 🌟