| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if 7 is a quadratic residue modulo 19 using Gauss's Lemma, we need to follow these steps: 1. List the least positive residues of the first (p-1)/2 multiples of a, where p is the prime modulus and a is the number we are checking if it is a quadratic residue. In this case, p = 19 and a = 7. So, we need to list the least positive residues of the first (19-1)/2 = 9 multiples of 7 modulo 19. 7 * 1 ≡ 7 (mod 19) 7 * 2 ≡ 14 (mod 19) 7 * 3 ≡ 2 (mod 19) 7 * 4 ≡ 9 (mod 19) 7 * 5 ≡ 16 (mod 19) 7 * 6 ≡ 4 (mod 19) 7 * 7 ≡ 11 (mod 19) 7 * 8 ≡ 18 (mod 19) 7 * 9 ≡ 6 (mod 19) 2. Count the number of least positive residues that are greater than p/2. In this case, p/2 = 9.5, so we need to count the number of least positive residues greater than 9.5. The least positive residues greater than 9.5 are 14, 16, 11, and 18. So, there are 4 such residues. 3. Calculate the Legendre symbol (a/p) using the formula: (a/p) = (-1)^S, where S is the number of least positive residues greater than p/2. In this case, S = 4, so (7/19) = (-1)^4 = 1. 4. If the Legendre symbol is 1, then a is a quadratic residue modulo p. If the Legendre symbol is -1, then a is not a quadratic residue modulo p. Since (7/19) = 1, 7 is a quadratic residue modulo 19. Solution: Yes, 7 is a quadratic residue modulo 19. |
True |
| deepseekr1 | To determine if 7 is a quadratic residue modulo 19 using Gauss's Lemma, we follow these steps: 1. **Gauss's Lemma**: For an odd prime \( p \) and an integer \( a \) not divisible by \( p \), the Legendre symbol \( \left( \frac{a}{p} \right) \) is equal to \( (-1)^n \), where \( n \) is the number of least residues of the multiples \( a, 2a, 3a, \ldots, \frac{p-1}{2}a \) that are greater than \( \frac{p}{2} \). 2. **Parameters**: Here, \( p = 19 \) and \( a = 7 \). We need to consider the multiples \( 7 \times 1, 7 \times 2, \ldots, 7 \times 9 \) (since \( \frac{19-1}{2} = 9 \)) and count how many of these residues are greater than \( 9.5 \). 3. **Calculate Residues**: - \( 7 \times 1 \equiv 7 \mod 19 \) (7 ≤ 9.5) - \( 7 \times 2 \equiv 14 \mod 19 \) (14 > 9.5) → count 1 - \( 7 \times 3 \equiv 21 \equiv 2 \mod 19 \) (2 ≤ 9.5) - \( 7 \times 4 \equiv 28 \equiv 9 \mod 19 \) (9 ≤ 9.5) - \( 7 \times 5 \equiv 35 \equiv 16 \mod 19 \) (16 > 9.5) → count 2 - \( 7 \times 6 \equiv 42 \equiv 4 \mod 19 \) (4 ≤ 9.5) - \( 7 \times 7 \equiv 49 \equiv 11 \mod 19 \) (11 > 9.5) → count 3 - \( 7 \times 8 \equiv 56 \equiv 18 \mod 19 \) (18 > 9.5) → count 4 - \( 7 \times 9 \equiv 63 \equiv 6 \mod 19 \) (6 ≤ 9.5) 4. **Result**: The number of residues greater than 9.5 is \( n = 4 \). According to Gauss's Lemma, \( \left( \frac{7}{19} \right) = (-1)^4 = 1 \), indicating that 7 is a quadratic residue modulo 19. 5. **Verification**: Using Euler's criterion, we compute \( 7^9 \mod 19 \): - \( 7^2 \equiv 11 \mod 19 \) - \( 7^4 \equiv 11^2 \equiv 7 \mod 19 \) - \( 7^8 \equiv 7^2 \equiv 11 \mod 19 \) - \( 7^9 \equiv 11 \times 7 \equiv 77 \equiv 1 \mod 19 \) This confirms \( \left( \frac{7}{19} \right) = 1 \). Additionally, \( 8^2 \equiv 64 \equiv 7 \mod 19 \) provides a concrete solution. Thus, the final answer is \(\boxed{Yes}\). |
True |
| gpto1 | To determine whether 7 is a quadratic residue modulo 19 using **Gauss's Lemma**, we'll follow these steps: **Gauss's Lemma**: Let \( p \) be an odd prime, and \( a \) an integer not divisible by \( p \). Let \( n \) be the number of integers \( k \) in \( \{1, 2, ..., \frac{p-1}{2}\} \) such that the least positive residue of \( ak \) modulo \( p \) is greater than \( \frac{p}{2} \). Then: \[ \left( \frac{a}{p} \right) = (-1)^n \] where \( \left( \frac{a}{p} \right) \) is the **Legendre symbol**, which equals 1 if \( a \) is a quadratic residue modulo \( p \), and -1 otherwise. **Applying Gauss's Lemma to \( a = 7 \) and \( p = 19 \)**: 1. **Calculate the Least Positive Residues**: Compute \( 7k \mod 19 \) for \( k = 1 \) to \( 9 \) (since \( \frac{19 - 1}{2} = 9 \)). | \( k \) | \( 7k \) | \( 7k \mod 19 \) | Comparison to \( \frac{19}{2} = 9.5 \) | |---------|---------|------------------|----------------------------------------| | 1 | 7 | 7 | \( 7 \leq 9.5 \) (Not counted) | | 2 | 14 | 14 | \( 14 > 9.5 \) (Counted, \( n = 1 \)) | | 3 | 21 | 2 | \( 2 \leq 9.5 \) (Not counted) | | 4 | 28 | 9 | \( 9 \leq 9.5 \) (Not counted) | | 5 | 35 | 16 | \( 16 > 9.5 \) (Counted, \( n = 2 \)) | | 6 | 42 | 4 | \( 4 \leq 9.5 \) (Not counted) | | 7 | 49 | 11 | \( 11 > 9.5 \) (Counted, \( n = 3 \)) | | 8 | 56 | 18 | \( 18 > 9.5 \) (Counted, \( n = 4 \)) | | 9 | 63 | 6 | \( 6 \leq 9.5 \) (Not counted) | 2. **Count the Number \( n \) of Residues Greater Than \( \frac{p}{2} \)**: From the table, the integers \( k \) where \( 7k \mod 19 > 9.5 \) are \( k = 2, 5, 7, \) and \( 8 \). Therefore, \( n = 4 \). 3. **Compute the Legendre Symbol**: Using Gauss's Lemma: \[ \left( \frac{7}{19} \right) = (-1)^n = (-1)^4 = 1 \] 4. **Conclusion**: Since the Legendre symbol \( \left( \frac{7}{19} \right) = 1 \), 7 is a quadratic residue modulo 19. **Answer**: Yes, 7 is a quadratic residue modulo 19 because \(\left( \frac{7}{19} \right) = (-1)^4 = 1\) using Gauss's Lemma. |
True |
| GPT-4o | To determine if 7 is a quadratic residue modulo 19 using Gauss's Lemma, we need to evaluate the Legendre symbol \(\left(\frac{7}{19}\right)\). According to Gauss's Lemma, the Legendre symbol \(\left(\frac{a}{p}\right)\) for an odd prime \(p\) and integer \(a\) is given by: \[ \left(\frac{a}{p}\right) = (-1)^n \] where \(n\) is the number of integers \(k\) in the set \(\{1, 2, \ldots, \frac{p-1}{2}\}\) such that the least positive residue of \(ak \mod p\) is greater than \(\frac{p}{2}\). For \(a = 7\) and \(p = 19\), we need to consider the set \(\{1, 2, \ldots, 9\}\) (since \(\frac{19-1}{2} = 9\)) and count how many values of \(k\) in this set make \(7k \mod 19\) greater than \(\frac{19}{2} = 9.5\). Let's compute \(7k \mod 19\) for \(k = 1, 2, \ldots, 9\): 1. \(7 \cdot 1 \mod 19 = 7\) 2. \(7 \cdot 2 \mod 19 = 14\) 3. \(7 \cdot 3 \mod 19 = 21 \equiv 2 \mod 19\) 4. \(7 \cdot 4 \mod 19 = 28 \equiv 9 \mod 19\) 5. \(7 \cdot 5 \mod 19 = 35 \equiv 16 \mod 19\) 6. \(7 \cdot 6 \mod 19 = 42 \equiv 4 \mod 19\) 7. \(7 \cdot 7 \mod 19 = 49 \equiv 11 \mod 19\) 8. \(7 \cdot 8 \mod 19 = 56 \equiv 18 \mod 19\) 9. \(7 \cdot 9 \mod 19 = 63 \equiv 5 \mod 19\) Next, we identify those results that are greater than 9.5: - \(14\) - \(16\) - \(11\) - \(18\) There are 4 values in the set \(\{14, 16, 11, 18\}\) that are greater than 9.5. So, \(n = 4\). Using Gauss's Lemma: \[ \left(\frac{7}{19}\right) = (-1)^n = (-1)^4 = 1 \] Since \(\left(\frac{7}{19}\right) = 1\), 7 is a quadratic residue modulo 19. |
True |
| llama3.1-70B-Instruct | To determine if 7 is a quadratic residue modulo 19 using Gauss's Lemma, we need to follow these steps: 1. **Understand Gauss's Lemma**: Gauss's Lemma states that for any odd prime \(p\) and any integer \(a\) not divisible by \(p\), the number of integers in the set \(\{a, 2a, 3a, \ldots, \frac{p-1}{2}a\}\) whose least residue modulo \(p\) is negative is even if and only if \(a\) is a quadratic residue modulo \(p\). 2. **Apply Gauss's Lemma to the problem**: Here, \(p = 19\) and \(a = 7\). We need to calculate the set \(\{7, 2 \times 7, 3 \times 7, \ldots, \frac{19-1}{2} \times 7\}\) modulo 19 and count how many of these have a negative least residue. 3. **Calculate the set**: The set is \(\{7, 14, 21, 28, 35, 42, 49, 56, 63\}\). Now, we reduce each element modulo 19: - \(7 \mod 19 = 7\) - \(14 \mod 19 = 14\) - \(21 \mod 19 = 2\) - \(28 \mod 19 = 9\) - \(35 \mod 19 = 16\) - \(42 \mod 19 = 4\) - \(49 \mod 19 = 11\) - \(56 \mod 19 = 18\) - \(63 \mod 19 = 6\) 4. **Count the negative residues**: We need to count how many of these residues are negative. However, since we're working with modulo, we consider residues that are greater than \(\frac{p}{2}\) as negative for the purpose of this calculation. Thus, we look for residues greater than 9 (since \(\frac{19}{2} = 9.5\)): - \(14\), \(16\), and \(18\) are greater than 9. 5. **Conclusion**: There are 3 residues that are considered "negative" by the criteria of Gauss's Lemma. Since 3 is an odd number, according to Gauss's Lemma, 7 is not a quadratic residue modulo 19. |
True |
| llama3.1-405B-Instruct | To determine if 7 is a quadratic residue modulo 19 using Gauss's Lemma, we need to calculate the Legendre symbol (7/19). Gauss's Lemma states that for any odd prime p and any integer a not divisible by p, the Legendre symbol (a/p) is equal to (-1)^n, where n is the number of integers in the set {a, 2a, 3a, ..., (p-1)a/2} that are congruent to 1 modulo p. In this case, we have a = 7 and p = 19. We need to calculate the number of integers in the set {7, 14, 21, ..., 9*7} that are congruent to 1 modulo 19. We can simplify the calculation by noting that 7 is congruent to -12 modulo 19, so we can replace 7 with -12 in the calculation. The set {7, 14, 21, ..., 9*7} is equivalent to the set {-12, -24, -36, ..., -108}. We can simplify this set by adding 19 to each element, which gives us the set {7, 5, 3, ..., 11}. Now, we can count the number of elements in this set that are congruent to 1 modulo 19. We find that there are 6 such elements: 7, 5, 3, 11, 9, and 1. Therefore, the Legendre symbol (7/19) is equal to (-1)^6, which is equal to 1. This means that 7 is a quadratic residue modulo 19. |
True |
| DeepSeek-Coder-V2 | To determine if 7 is a quadratic residue modulo 19 using Gauss's Lemma, we need to follow these steps: 1. **Identify the relevant values:** - \( p = 19 \) - \( a = 7 \) 2. **Calculate the least positive residues of \( a \cdot k \mod p \) for \( k = 1, 2, \ldots, \frac{p-1}{2} \):** - \( \frac{p-1}{2} = \frac{19-1}{2} = 9 \) - We need to compute \( 7 \cdot k \mod 19 \) for \( k = 1, 2, \ldots, 9 \). 3. **Compute the least positive residues:** - \( 7 \cdot 1 \mod 19 = 7 \) - \( 7 \cdot 2 \mod 19 = 14 \) - \( 7 \cdot 3 \mod 19 = 21 \mod 19 = 2 \) - \( 7 \cdot 4 \mod 19 = 28 \mod 19 = 9 \) - \( 7 \cdot 5 \mod 19 = 35 \mod 19 = 16 \) - \( 7 \cdot 6 \mod 19 = 42 \mod 19 = 4 \) - \( 7 \cdot 7 \mod 19 = 49 \mod 19 = 11 \) - \( 7 \cdot 8 \mod 19 = 56 \mod 19 = 18 \) - \( 7 \cdot 9 \mod 19 = 63 \mod 19 = 6 \) 4. **Count the number of residues that are greater than \( \frac{p}{2} \):** - \( \frac{p}{2} = \frac{19}{2} = 9.5 \) - The residues greater than 9.5 are: 11, 14, 16, 18 - There are 4 such residues. 5. **Apply Gauss's Lemma:** - Gauss's Lemma states that \( a \) is a quadratic residue modulo \( p \) if and only if the number of residues greater than \( \frac{p}{2} \) is even. - Here, the number of residues greater than 9.5 is 4, which is even. Therefore, 7 is a quadratic residue modulo 19. |
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| Qwen2-72B-Instruct | To determine if 7 is a quadratic residue modulo 19 using Gauss's Lemma, we need to follow these steps: 1. List the least positive residues of the first (p-1)/2 multiples of a, where p is the prime modulus and a is the number we are checking if it is a quadratic residue. In this case, p = 19 and a = 7. So, we need to list the least positive residues of the first (19-1)/2 = 9 multiples of 7 modulo 19. 7 * 1 ≡ 7 (mod 19) 7 * 2 ≡ 14 (mod 19) 7 * 3 ≡ 2 (mod 19) 7 * 4 ≡ 9 (mod 19) 7 * 5 ≡ 16 (mod 19) 7 * 6 ≡ 4 (mod 19) 7 * 7 ≡ 11 (mod 19) 7 * 8 ≡ 18 (mod 19) 7 * 9 ≡ 6 (mod 19) 2. Count the number of least positive residues that are greater than p/2. In this case, p/2 = 9.5, so we need to count the number of least positive residues greater than 9.5. The least positive residues greater than 9.5 are 14, 16, 11, and 18. So, there are 4 such residues. 3. Calculate the Legendre symbol (a/p) using the formula: (a/p) = (-1)^S, where S is the number of least positive residues greater than p/2. In this case, S = 4, so (7/19) = (-1)^4 = 1. 4. If the Legendre symbol is 1, then a is a quadratic residue modulo p. If the Legendre symbol is -1, then a is not a quadratic residue modulo p. Since (7/19) = 1, 7 is a quadratic residue modulo 19. Solution: Yes, 7 is a quadratic residue modulo 19. |
True |
| baichuan4 | To determine if 7 is a quadratic residue modulo 19 using Gauss's Lemma, we need to find an integer x such that x^2 ≡ 7 (mod 19). Gauss's Lemma states that if p is an odd prime and a is an integer not divisible by p, then a is a quadratic residue modulo p if and only if the product of its nonzero digits in its base p representation is congruent to a modulo p. First, we convert 7 into its base 19 representation. 7 in base 19 is 12 (since 19 > 7, we only need one digit). Next, we apply Gauss's Lemma by computing the product of the digits and reducing it modulo 19: 12 (mod 19) = 12 Since 12^2 ≡ 144 ≡ 7 (mod 19), we have found an integer (12) such that its square is congruent to 7 modulo 19. Therefore, 7 is a quadratic residue modulo 19. |
True |
| doubao | Gauss's Lemma states that for an odd prime p and an integer a not divisible by p, the Legendre symbol (a/p) is equal to (-1)^m, where m is the number of integers in the set {a, 2a, 3a,...,(p-1)a/2} that are greater than p/2. For a = 7 and p = 19: The set {7, 14, 21, 28, 35, 42, 49, 56, 63, 70} (mod 19) is {7, 14, 2, 9, 16, 3, 10, 17, 4, 11}. There are 5 numbers in this set that are greater than 19/2 = 9.5, so m = 5. Then (-1)^5 = -1. So, 7 is not a quadratic residue modulo 19. |
True |
| minimax | Gauss's Lemma provides a method to determine whether a number \( a \) is a quadratic residue modulo a prime \( p \). According to Gauss's Lemma, \( a \) is a quadratic residue modulo \( p \) if and only if the number of integers in the set \( \{a, 2a, 3a, ..., \frac{p-1}{2}a\} \) that are greater than \( \frac{p}{2} \) is even. Let's apply Gauss's Lemma to determine if 7 is a quadratic residue modulo 19. We need to consider the multiples of 7 modulo 19, up to \( \frac{19-1}{2} = 9 \) multiples: 1. \( 1 \cdot 7 \equiv 7 \mod 19 \) 2. \( 2 \cdot 7 \equiv 14 \mod 19 \) 3. \( 3 \cdot 7 \equiv 21 \equiv 2 \mod 19 \) 4. \( 4 \cdot 7 \equiv 28 \equiv 9 \mod 19 \) 5. \( 5 \cdot 7 \equiv 35 \equiv 16 \mod 19 \) 6. \( 6 \cdot 7 \equiv 42 \equiv 4 \mod 19 \) 7. \( 7 \cdot 7 \equiv 49 \equiv 11 \mod 19 \) 8. \( 8 \cdot 7 \equiv 56 \equiv 18 \mod 19 \) 9. \( 9 \cdot 7 \equiv 63 \equiv 6 \mod 19 \) Now, we need to count how many of these multiples are greater than \( \frac{19}{2} = 9.5 \). Since we are working modulo 19, we can consider the numbers greater than 9 as well: - 7 is less than 9.5. - 14 is less than 9.5. - 2 is less than 9.5. - 9 is less than 9.5. - 16 is greater than 9.5. - 4 is less than 9.5. - 11 is greater than 9.5. - 18 is greater than 9.5. - 6 is less than 9.5. Out of these 9 multiples, 3 are greater than \( \frac{19}{2} \) (16, 11, and 18). Since this number is odd, according to Gauss's Lemma, 7 is not a quadratic residue modulo 19. |
True |
| kimi | To determine if 7 is a quadratic residue modulo 19, we can use Gauss's Lemma. Gauss's Lemma states that a number \( a \) is a quadratic residue modulo \( p \) if and only if there exists an integer \( x \) such that \( x^2 \equiv a \pmod{p} \). For \( p = 19 \), we need to check if there exists an integer \( x \) such that \( x^2 \equiv 7 \pmod{19} \). However, Gauss's Lemma is more directly applied to check if a number is a quadratic non-residue modulo \( p \). To use Gauss's Lemma for this purpose, we need to find a number \( a \) such that \( a^{(p-1)/2} \not\equiv 1 \pmod{p} \). If such an \( a \) exists, then \( a \) is a quadratic non-residue modulo \( p \). For \( p = 19 \), we would check \( a^{9} \pmod{19} \) for \( a = 7 \). Let's check if 7 is a quadratic residue modulo 19: We need to find \( x \) such that \( x^2 \equiv 7 \pmod{19} \). We can do this by checking all possible values of \( x \) from 1 to 9 (since \( x \) and \( 19-x \) would give the same quadratic residue). If we find such an \( x \), then 7 is a quadratic residue modulo 19. Let's check: \( 1^2 \equiv 1 \pmod{19} \) \( 2^2 \equiv 4 \pmod{19} \) \( 3^2 \equiv 9 \pmod{19} \) \( 4^2 \equiv 16 \pmod{19} \) \( 5^2 \equiv 25 \equiv 6 \pmod{19} \) \( 6^2 \equiv 36 \equiv 17 \pmod{19} \) \( 7^2 \equiv 49 \equiv 11 \pmod{19} \) \( 8^2 \equiv 64 \equiv 7 \pmod{19} \) \( 9^2 \equiv 81 \equiv 4 \pmod{19} \) We find that \( 8^2 \equiv 64 \equiv 7 \pmod{19} \), so there is an integer \( x = 8 \) such that \( x^2 \equiv 7 \pmod{19} \). Therefore, 7 is a quadratic residue modulo 19. |
True |
| wenxinyiyan | To determine if 7 is a quadratic residue modulo 19, we can use Gauss's Lemma. Gauss's Lemma states that for an odd prime p and an integer a such that p does not divide a, the Legendre symbol (a/p) is equal to (-1)^μ, where μ is the number of integers in the set {a, 2a, 3a, ..., (p-1)/2 * a} whose least positive residues modulo p are greater than p/2. In our case, p = 19 and a = 7. Let's calculate the set {7, 2*7, 3*7, ..., 9*7} modulo 19 and count how many least positive residues are greater than 19/2 = 9.5, i.e., greater than 9. 1. 7 ≡ 7 (mod 19) 2. 2*7 = 14 ≡ 14 (mod 19) 3. 3*7 = 21 ≡ 2 (mod 19) 4. 4*7 = 28 ≡ 9 (mod 19) 5. 5*7 = 35 ≡ 16 (mod 19) 6. 6*7 = 42 ≡ 4 (mod 19) 7. 7*7 = 49 ≡ 11 (mod 19) 8. 8*7 = 56 ≡ 18 (mod 19) 9. 9*7 = 63 ≡ 6 (mod 19) The least positive residues that are greater than 9 are {14, 16, 11, 18}. There are 4 of them. According to Gauss's Lemma, the Legendre symbol (7/19) is equal to (-1)^4 = 1. Therefore, 7 is a quadratic residue modulo 19. |
True |
| zhipu | To determine if 7 is a quadratic residue modulo 19 using Gauss's Lemma, we will follow these steps: ### Step 1: Understand Gauss's Lemma Gauss's Lemma states that if \( p \) is an odd prime and \( a \) is an integer not divisible by \( p \), then \( a \) is a quadratic residue modulo \( p \) if and only if the number of positive integers less than \( \frac{p-1}{2} \) that are congruent to \( a \cdot k \mod p \) (for \( k = 1, 2, \ldots, \frac{p-1}{2} \)) and are greater than \( \frac{p}{2} \) is even. ### Step 2: Apply Gauss's Lemma to the given problem Here, \( p = 19 \) and \( a = 7 \). 1. Calculate \( \frac{p-1}{2} \): \[ \frac{19-1}{2} = 9 \] 2. List the integers \( 1, 2, \ldots, 9 \). 3. Compute \( 7k \mod 19 \) for \( k = 1, 2, \ldots, 9 \): \[ \begin{align*} 7 \cdot 1 &\equiv 7 \mod 19 \\ 7 \cdot 2 &\equiv 14 \mod 19 \\ 7 \cdot 3 &\equiv 21 \equiv 2 \mod 19 \\ 7 \cdot 4 &\equiv 28 \equiv 9 \mod 19 \\ 7 \cdot 5 &\equiv 35 \equiv 16 \mod 19 \\ 7 \cdot 6 &\equiv 42 \equiv 3 \mod 19 \\ 7 \cdot 7 &\equiv 49 \equiv 11 \mod 19 \\ 7 \cdot 8 &\equiv 56 \equiv 18 \mod 19 \\ 7 \cdot 9 &\equiv 63 \equiv 5 \mod 19 \\ \end{align*} \] 4. Identify which of these values are greater than \( \frac{19}{2} = 9.5 \): \[ 14, 16, 11, 18 \] 5. Count the number of such values: There are 4 values: \( 14, 16, 11, 18 \). ### Step 3: Determine if 7 is a quadratic residue modulo 19 Since the number of values greater than \( \frac{19}{2} \) is 4, which is even, by Gauss's Lemma, 7 is a quadratic residue modulo 19. ### Conclusion Yes, 7 is a quadratic residue modulo 19. |
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| gpto1 | To determine whether 7 is a quadratic residue modulo 19 using **Gauss's Lemma**, we'll follow these steps: **Gauss's Lemma**: Let \( p \) be an odd prime, and \( a \) an integer not divisible by \( p \). Let \( n \) be the number of integers \( k \) in \( \{1, 2, ..., \frac{p-1}{2}\} \) such that the least positive residue of \( ak \) modulo \( p \) is greater than \( \frac{p}{2} \). Then: \[ \left( \frac{a}{p} \right) = (-1)^n \] where \( \left( \frac{a}{p} \right) \) is the **Legendre symbol**, which equals 1 if \( a \) is a quadratic residue modulo \( p \), and -1 otherwise. **Applying Gauss's Lemma to \( a = 7 \) and \( p = 19 \)**: 1. **Calculate the Least Positive Residues**: Compute \( 7k \mod 19 \) for \( k = 1 \) to \( 9 \) (since \( \frac{19 - 1}{2} = 9 \)). | \( k \) | \( 7k \) | \( 7k \mod 19 \) | Comparison to \( \frac{19}{2} = 9.5 \) | |---------|---------|------------------|----------------------------------------| | 1 | 7 | 7 | \( 7 \leq 9.5 \) (Not counted) | | 2 | 14 | 14 | \( 14 > 9.5 \) (Counted, \( n = 1 \)) | | 3 | 21 | 2 | \( 2 \leq 9.5 \) (Not counted) | | 4 | 28 | 9 | \( 9 \leq 9.5 \) (Not counted) | | 5 | 35 | 16 | \( 16 > 9.5 \) (Counted, \( n = 2 \)) | | 6 | 42 | 4 | \( 4 \leq 9.5 \) (Not counted) | | 7 | 49 | 11 | \( 11 > 9.5 \) (Counted, \( n = 3 \)) | | 8 | 56 | 18 | \( 18 > 9.5 \) (Counted, \( n = 4 \)) | | 9 | 63 | 6 | \( 6 \leq 9.5 \) (Not counted) | 2. **Count the Number \( n \) of Residues Greater Than \( \frac{p}{2} \)**: From the table, the integers \( k \) where \( 7k \mod 19 > 9.5 \) are \( k = 2, 5, 7, \) and \( 8 \). Therefore, \( n = 4 \). 3. **Compute the Legendre Symbol**: Using Gauss's Lemma: \[ \left( \frac{7}{19} \right) = (-1)^n = (-1)^4 = 1 \] 4. **Conclusion**: Since the Legendre symbol \( \left( \frac{7}{19} \right) = 1 \), 7 is a quadratic residue modulo 19. **Answer**: Yes, 7 is a quadratic residue modulo 19 because \(\left( \frac{7}{19} \right) = (-1)^4 = 1\) using Gauss's Lemma. |
True |
| deepseekv3 | To determine whether 7 is a quadratic residue modulo 19 using **Gauss's Lemma**, follow these steps: --- ### **Gauss's Lemma Statement**: Let \( p \) be an odd prime and \( a \) be an integer such that \( p \nmid a \). Consider the sequence: \[ a, 2a, 3a, \dots, \left(\frac{p-1}{2}\right)a \] Let \( \mu \) be the number of elements in this sequence whose least positive residues modulo \( p \) are greater than \( \frac{p}{2} \). Then: \[ \left(\frac{a}{p}\right) = (-1)^\mu, \] where \( \left(\frac{a}{p}\right) \) is the Legendre symbol. If \( \left(\frac{a}{p}\right) = 1 \), \( a \) is a quadratic residue modulo \( p \); otherwise, it is not. --- ### **Step 1: Set up the problem** We are given \( a = 7 \) and \( p = 19 \). Compute the sequence: \[ 7, 2 \cdot 7, 3 \cdot 7, \dots, \left(\frac{19-1}{2}\right) \cdot 7 \] This becomes: \[ 7, 14, 21, 28, 35, 42, 49, 56, 63 \] --- ### **Step 2: Compute residues modulo 19** Reduce each term modulo 19: \[ 7 \equiv 7 \pmod{19} \\ 14 \equiv 14 \pmod{19} \\ 21 \equiv 2 \pmod{19} \\ 28 \equiv 9 \pmod{19} \\ 35 \equiv 16 \pmod{19} \\ 42 \equiv 4 \pmod{19} \\ 49 \equiv 11 \pmod{19} \\ 56 \equiv 18 \pmod{19} \\ 63 \equiv 6 \pmod{19} \] --- ### **Step 3: Count residues greater than \( \frac{19}{2} = 9.5 \)** From the residues \( 7, 14, 2, 9, 16, 4, 11, 18, 6 \), the values greater than \( 9.5 \) are: \[ 14, 16, 11, 18 \] Thus, \( \mu = 4 \). --- ### **Step 4: Compute the Legendre symbol** Using Gauss's Lemma: \[ \left(\frac{7}{19}\right) = (-1)^4 = 1 \] --- ### **Conclusion**: Since \( \left(\frac{7}{19}\right) = 1 \), \( 7 \) is a **quadratic residue modulo 19**. \[ \boxed{1} \] |
True |