Let p = 3 and q = 7. Which statement correctly applies the law of quadratic reciprocity?
AThe law gives (3/7) = (7/3), because reciprocity always makes the two Legendre symbols equal
B(3/7) = −(7/3), because both primes are ≡ 3 (mod 4), making the exponent odd
C(3/7)(7/3) = 1, because p and q are both odd primes
DThe law does not apply because 3 and 7 are small enough to evaluate directly
The law says (p/q)(q/p) = (−1)^{(p−1)/2 · (q−1)/2}. Here (3−1)/2 = 1 and (7−1)/2 = 3, so the exponent is 3, giving (−1)^3 = −1. Therefore (3/7) and (7/3) have opposite signs. The classic misconception is option A — that reciprocity makes the two symbols equal. It does not; it relates their *product* to ±1.
Question 2 Multiple Choice
You want to compute (17/101). Since 17 ≡ 1 (mod 4), what does quadratic reciprocity allow you to conclude?
A(17/101) = −(101/17), so they have opposite signs
B(17/101) = (101/17), so you can replace the computation with the simpler (101/17)
CThe law provides no simplification — you must compute (17/101) directly
D(17/101) = 0 because 101 does not divide 17
When p ≡ 1 (mod 4), (p−1)/2 is even, so the exponent (p−1)/2 · (q−1)/2 is even regardless of q, giving (p/q)(q/p) = 1. Therefore (17/101) = (101/17). Now 101 = 5·17 + 16, so (101/17) = (16/17) = (4²/17) = 1. This reduction — analogous to the Euclidean algorithm — is the computational power of the law.
Question 3 True / False
If p ≡ 1 (mod 4), then (p/q) = (q/p) for any distinct odd prime q.
TTrue
FFalse
Answer: True
When p ≡ 1 (mod 4), (p−1)/2 is even. The exponent in (−1)^{(p−1)/2 · (q−1)/2} is therefore always even (an even number times anything is even), giving product 1. So (p/q)(q/p) = 1, meaning (p/q) = (q/p). The signs only flip when BOTH p and q are ≡ 3 (mod 4) — only then is the exponent odd.
Question 4 True / False
The law of quadratic reciprocity states that (p/q) = (q/p) for most distinct odd primes p and q.
TTrue
FFalse
Answer: False
This is the most common misstatement of the law. The law says (p/q)(q/p) = (−1)^{(p−1)/2 · (q−1)/2}, not that (p/q) = (q/p). When both p ≡ q ≡ 3 (mod 4), the product equals −1, so the two Legendre symbols are *opposite* in sign. For example, (3/7) = −1 and (7/3) = 1, confirming they differ.
Question 5 Short Answer
The law gives (p/q)(q/p) = ±1. When does the product equal −1, and what determines the sign?
Think about your answer, then reveal below.
Model answer: The product equals −1 when both p ≡ 3 (mod 4) and q ≡ 3 (mod 4). The exponent (p−1)/2 · (q−1)/2 is odd only when both factors are odd — and (p−1)/2 is odd exactly when p ≡ 3 (mod 4). So the sign is −1 iff both primes are 3 mod 4; otherwise it is +1.
The key insight is that the sign depends not on the primes themselves but solely on their residues mod 4. When p ≡ 1 (mod 4), the factor (p−1)/2 is even, killing any sign contribution from p. The flip only occurs when both primes contribute an odd factor — i.e., both are ≡ 3 (mod 4). This elegant structure is why the law is both surprising (connecting solvability of two unrelated congruences) and computationally useful.