Find the number of solution pairs for an equation in a specific algebraic ring structure | Step-by-Step Solution

TinyProf's Solution Guide 🎓

What We're Solving:

We need to find how many pairs of elements (u,v) in the ring Z_m[x]/(x²-δ) multiply together to give n, where m = 2^q and δ is a square-free quadratic non-residue modulo m.

The Approach:

The ring is a "number system" that's like the integers mod m, but with an extra "imaginary" component involving √δ. We're essentially counting solutions in a 2-dimensional structure over Z_m. The key insight is to use the norm map and leverage properties of quadratic forms over rings of the form Z_{2^q}.

Step-by-Step Solution:

Step 1: Understand the Ring Structure

• Elements in Z_m[x]/(x²-δ) look like a + bx where a,b ∈ Z_m
• We can think of x as "√δ", so elements are a + b√δ
• Multiplication follows: (a + b√δ)(c + d√δ) = (ac + bdδ) + (ad + bc)√δ

Step 2: Set Up the Equation

• Let u = a + b√δ and v = c + d√δ
• Let n = n₀ + n₁√δ (our target element)
• Then uv = n gives us the system:

- ac + bdδ = n₀ - ad + bc = n₁

Step 3: Use the Norm Map

• Define N(a + b√δ) = a² - b²δ (the norm)
• Key property: N(uv) = N(u)N(v)
• So we need N(u)N(v) = N(n) = n₀² - n₁²δ

Step 4: Apply 2-adic Theory Since m = 2^q and δ is a quadratic non-residue:

• The ring Z_m[x]/(x²-δ) has exactly m² elements
• For each fixed u ≠ 0, the equation uv = n has a unique solution v (when it exists)
• The number of solutions depends on whether n is a unit in the ring

Step 5: Count Solutions

• If n = 0: We get m² - 1 solutions (any u,v where uv = 0 but not both zero, plus (0,0))
• If n is a unit: We get exactly m solutions
• If n is a zero divisor but ≠ 0: We get m solutions

Step 6: Determine When n is a Unit In Z_{2^q}[x]/(x²-δ), an element n₀ + n₁√δ is a unit if and only if gcd(n₀² - n₁²δ, 2^q) = 1.

The Answer:

The number of solutions to uv = n in Z_{2^q}[x]/(x²-δ) is:

• 2^(2q) solutions if n = 0
• 2^q solutions if n ≠ 0

This uniform count for all non-zero n occurs because in the 2-adic case with a quadratic non-residue, every non-zero element has the same "solvability pattern."

Memory Tip: 🧠

Remember "Square the base for zero, keep the base for non-zero!" When m = 2^q, zero gives you m² solutions, and any non-zero element gives you exactly m solutions. This pattern reflects the beautiful structure of quadratic extensions over 2-adic rings! 🌟