| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if 10 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, modulo p. In this case, p = 19 and a = 10. So, we need to find the least positive residues of the first (19-1)/2 = 9 multiples of 10 modulo 19. 2. Count the number of least positive residues that are greater than p/2. If the number of such residues is even, then a is a quadratic residue modulo p. If the number of such residues is odd, then a is not a quadratic residue modulo p. Let's calculate the least positive residues of the first 9 multiples of 10 modulo 19: 1. 10 * 1 ≡ 10 (mod 19) 2. 10 * 2 ≡ 20 ≡ 1 (mod 19) 3. 10 * 3 ≡ 30 ≡ 11 (mod 19) 4. 10 * 4 ≡ 40 ≡ 2 (mod 19) 5. 10 * 5 ≡ 50 ≡ 12 (mod 19) 6. 10 * 6 ≡ 60 ≡ 3 (mod 19) 7. 10 * 7 ≡ 70 ≡ 14 (mod 19) 8. 10 * 8 ≡ 80 ≡ 4 (mod 19) 9. 10 * 9 ≡ 90 ≡ 16 (mod 19) Now, let's count the number of least positive residues that are greater than p/2 = 19/2 = 9.5: 1. 10 > 9.5 2. 11 > 9.5 3. 12 > 9.5 4. 14 > 9.5 5. 16 > 9.5 There are 5 least positive residues greater than 9.5, which is an odd number. Therefore, according to Gauss's Lemma, 10 is not a quadratic residue modulo 19. |
False |
| deepseekr1 | To determine if 10 is a quadratic residue modulo 19 using Gauss's Lemma, we need to consider the least residues of the multiples of 10 modulo 19 for the integers from 1 to 9 (since \((19-1)/2 = 9\)). We then count how many of these residues are greater than \(9.5\). Calculating the residues: - \(1 \times 10 \equiv 10 \mod 19\) (greater than 9.5) - \(2 \times 10 \equiv 20 \equiv 1 \mod 19\) (not greater than 9.5) - \(3 \times 10 \equiv 30 \equiv 11 \mod 19\) (greater than 9.5) - \(4 \times 10 \equiv 40 \equiv 2 \mod 19\) (not greater than 9.5) - \(5 \times 10 \equiv 50 \equiv 12 \mod 19\) (greater than 9.5) - \(6 \times 10 \equiv 60 \equiv 3 \mod 19\) (not greater than 9.5) - \(7 \times 10 \equiv 70 \equiv 13 \mod 19\) (greater than 9.5) - \(8 \times 10 \equiv 80 \equiv 4 \mod 19\) (not greater than 9.5) - \(9 \times 10 \equiv 90 \equiv 14 \mod 19\) (greater than 9.5) The residues greater than 9.5 are 10, 11, 12, 13, and 14, which totals 5 residues. Since 5 is odd, Gauss's Lemma tells us that 10 is a quadratic non-residue modulo 19. To verify, we use Euler's Criterion, calculating \(10^{(19-1)/2} \mod 19\): - \(10^2 \equiv 5 \mod 19\) - \(10^4 \equiv 5^2 \equiv 6 \mod 19\) - \(10^8 \equiv 6^2 \equiv 17 \mod 19\) - \(10^9 \equiv 17 \times 10 \equiv 170 \equiv -1 \mod 19\) Since \(10^9 \equiv -1 \mod 19\), Euler's Criterion confirms 10 is a quadratic non-residue modulo 19. \[ \boxed{No} \] |
False |
| gpto1 | To determine whether 10 is a quadratic residue modulo 19 using **Gauss's Lemma**, we'll follow these steps: **Gauss's Lemma Statement:** For an odd prime \( p \) and an integer \( a \) not divisible by \( p \), let \( n \) be the least positive residue of \( a \) modulo \( p \). Consider the set: \[ S = \{ n, 2n, 3n, \dotsc, \left( \frac{p-1}{2} \right) n \} \] Reduce each element of \( S \) modulo \( p \) to get their least positive residues. Let \( m \) be the number of these residues that are greater than \( \frac{p}{2} \). Then: \[ \left( \frac{a}{p} \right) = (-1)^m \] where \( \left( \frac{a}{p} \right) \) is the **Legendre symbol**. If \( \left( \frac{a}{p} \right) = 1 \), then \( a \) is a quadratic residue modulo \( p \); if \( \left( \frac{a}{p} \right) = -1 \), then \( a \) is a quadratic non-residue modulo \( p \). **Step-by-Step Solution:** 1. **Identify Parameters:** - \( p = 19 \) (an odd prime) - \( a = 10 \) - \( n = 10 \) (since \( a \mod p = 10 \)) 2. **Compute \( \frac{p - 1}{2} \):** \[ \frac{19 - 1}{2} = 9 \] So we need to consider multiples of \( n \) from \( 1 \) to \( 9 \). 3. **Construct the Set \( S \):** \[ S = \{ n, 2n, 3n, \dotsc, 9n \} = \{10, 20, 30, \dotsc, 90\} \] 4. **Compute Least Positive Residues Modulo \( p \):** Reduce each element in \( S \) modulo \( 19 \): \[ \begin{align*} s_1 &= 10 \mod 19 = 10 \\ s_2 &= 20 \mod 19 = 1 \\ s_3 &= 30 \mod 19 = 11 \\ s_4 &= 40 \mod 19 = 2 \\ s_5 &= 50 \mod 19 = 12 \\ s_6 &= 60 \mod 19 = 3 \\ s_7 &= 70 \mod 19 = 13 \\ s_8 &= 80 \mod 19 = 4 \\ s_9 &= 90 \mod 19 = 14 \\ \end{align*} \] 5. **Determine Which Residues Are Greater Than \( \frac{p}{2} \):** Calculate \( \frac{p}{2} \): \[ \frac{19}{2} = 9.5 \] List the residues and mark those greater than 9.5: - \( s_1 = 10 \) (> 9.5) - \( s_2 = 1 \) (< 9.5) - \( s_3 = 11 \) (> 9.5) - \( s_4 = 2 \) (< 9.5) - \( s_5 = 12 \) (> 9.5) - \( s_6 = 3 \) (< 9.5) - \( s_7 = 13 \) (> 9.5) - \( s_8 = 4 \) (< 9.5) - \( s_9 = 14 \) (> 9.5) Residues greater than \( \frac{p}{2} \) are \( s_1, s_3, s_5, s_7, s_9 \). 6. **Count \( m \):** Number of residues greater than \( 9.5 \): \[ m = 5 \] 7. **Compute the Legendre Symbol Using Gauss's Lemma:** \[ \left( \frac{10}{19} \right) = (-1)^m = (-1)^5 = -1 \] 8. **Conclusion:** Since \( \left( \frac{10}{19} \right) = -1 \), the number \( 10 \) is a **quadratic non-residue** modulo \( 19 \). **Answer:** **No, 10 is not a quadratic residue modulo 19; using Gauss's Lemma, we find that (10 | 19) = –1.** |
False |
| GPT-4o | To determine whether 10 is a quadratic residue modulo 19 using Gauss's Lemma, we follow these steps: ### Step 1: Definition and Setup Gauss's Lemma states that for an integer \( a \) and an odd prime \( p \), \( a \) is a quadratic residue modulo \( p \) if and only if the number of integers \( k \) in the set \(\{1, 2, ..., \frac{p-1}{2}\}\) for which the least positive residue of \( ak \) modulo \( p \) is greater than \( \frac{p}{2} \) is even. Here, \( a = 10 \) and \( p = 19 \). We need to check the least positive residues of \( 10k \) modulo 19 for \( k = 1, 2, \ldots, 9 \) (since \( \frac{19-1}{2} = 9 \)). ### Step 2: Calculate the Least Positive Residues We calculate \( 10k \mod 19 \) for \( k \) from 1 to 9: - For \( k = 1 \): \( 10 \cdot 1 = 10 \), and \( 10 \mod 19 = 10 \) - For \( k = 2 \): \( 10 \cdot 2 = 20 \), and \( 20 \mod 19 = 1 \) - For \( k = 3 \): \( 10 \cdot 3 = 30 \), and \( 30 \mod 19 = 11 \) - For \( k = 4 \): \( 10 \cdot 4 = 40 \), and \( 40 \mod 19 = 2 \) - For \( k = 5 \): \( 10 \cdot 5 = 50 \), and \( 50 \mod 19 = 12 \) - For \( k = 6 \): \( 10 \cdot 6 = 60 \), and \( 60 \mod 19 = 3 \) - For \( k = 7 \): \( 10 \cdot 7 = 70 \), and \( 70 \mod 19 = 13 \) - For \( k = 8 \): \( 10 \cdot 8 = 80 \), and \( 80 \mod 19 = 4 \) - For \( k = 9 \): \( 10 \cdot 9 = 90 \), and \( 90 \mod 19 = 14 \) ### Step 3: Identify Residues Greater Than \( \frac{19}{2} = 9.5 \) Now, we identify which of these residues are greater than 9.5: - \( 10 \) (greater than 9.5) - \( 11 \) (greater than 9.5) - \( 12 \) (greater than 9.5) - \( 13 \) (greater than 9.5) - \( 14 \) (greater than 9.5) There are 5 residues (10, 11, 12, 13, 14) greater than 9.5. ### Step 4: Apply Gauss's Lemma According to Gauss's Lemma, \( 10 \) is a quadratic residue modulo \( 19 \) if the number of residues greater than \( \frac{19}{2} \) is even. Since we found 5 such residues, which is odd, \( 10 \) is **not** a quadratic residue modulo \( 19 \). ### Conclusion Using Gauss's Lemma, we conclude that 10 is not a quadratic residue modulo 19. |
False |
| llama3.1-70B-Instruct | To determine if 10 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 \(\{a, 2a, 3a, \ldots, \frac{p-1}{2}a\}\) that are congruent to their negatives modulo \(p\) (i.e., \(ka \equiv -ka \mod p\)) is even if and only if \(a\) is a quadratic residue modulo \(p\). 2. **Apply Gauss's Lemma to the problem**: Here, \(a = 10\) and \(p = 19\). We need to examine the set \(\{10, 20, 30, \ldots, \frac{19-1}{2} \times 10\}\) modulo 19. Since \(\frac{19-1}{2} = 9\), we look at the set \(\{10, 20, 30, 40, 50, 60, 70, 80, 90\}\) modulo 19. 3. **Reduce each element modulo 19**: - \(10 \mod 19 = 10\) - \(20 \mod 19 = 1\) - \(30 \mod 19 = 12\) - \(40 \mod 19 = 2\) - \(50 \mod 19 = 13\) - \(60 \mod 19 = 3\) - \(70 \mod 19 = 14\) - \(80 \mod 19 = 4\) - \(90 \mod 19 = 15\) 4. **Determine the number of elements congruent to their negatives**: - \(10 \equiv -9 \mod 19\) - \(1 \equiv -18 \mod 19\) - \(12 \equiv -7 \mod 19\) - \(2 \equiv -17 \mod 19\) - \(13 \equiv -6 \mod 19\) - \(3 \equiv -16 \mod 19\) - \(14 \equiv -5 \mod 19\) - \(4 \equiv -15 \mod 19\) - \(15 \equiv -4 \mod 19\) There are 9 elements, and all of them are congruent to their negatives modulo 19, but since we are counting the number of elements that are congruent to their negatives in the set \(\{a, 2a, \ldots, \frac{p-1}{2}a\}\), we notice that the condition for applying Gauss's Lemma directly as stated initially might have been misinterpreted in the explanation. The correct application involves counting the number of elements that are in the "upper half" of the residues (i.e., greater than \(\frac{p-1}{2}\)) when reduced modulo \(p\), which would indicate a quadratic residue if the count is even. 5. **Correct Application of Gauss's Lemma**: - The correct count should consider how many of the original set \(\{10, 20, 30, \ldots, 90\}\) fall into the "upper half" when reduced modulo 19. However, the initial explanation mistakenly applied the concept directly without considering the actual process of counting elements in the upper half of residues modulo 19. Given the confusion in the explanation and the incorrect direct application of Gauss's Lemma as described, let's correct the approach: To correctly apply Gauss's Lemma, one should count the number of elements in the set \(\{10, 20, 30, \ldots, 90\}\) that, when reduced modulo 19, fall into the range \(\{10, 11, \ldots, 18\}\), which represents the "upper half" of residues modulo 19. However, the detailed step-by-step process provided initially does not accurately reflect this correct application. Given the mistake in the explanation process, the correct conclusion based on Gauss's Lemma cannot be accurately derived from the provided steps. Gauss's Lemma is a powerful tool for determining quadratic residues, but its application requires careful consideration of the specific conditions and correct counting of residues in the context of the problem at hand. |
False |
| llama3.1-405B-Instruct | To determine if 10 is a quadratic residue modulo 19 using Gauss's Lemma, we need to calculate the Legendre symbol (10/19). 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 = 10 and p = 19. We need to calculate the number of integers in the set {10, 20, 30, ..., 90} that are congruent to 1 modulo 19. We can simplify the calculation by noting that 10 is congruent to -9 modulo 19, so we can replace 10 with -9 in the set. We get: {-9, -18, -27, ..., -81} Now, we can calculate the number of integers in this set that are congruent to 1 modulo 19: * -9 is congruent to 10 modulo 19 (not 1) * -18 is congruent to 1 modulo 19 (yes) * -27 is congruent to 12 modulo 19 (not 1) * -36 is congruent to 3 modulo 19 (not 1) * -45 is congruent to 14 modulo 19 (not 1) * -54 is congruent to 5 modulo 19 (not 1) * -63 is congruent to 16 modulo 19 (not 1) * -72 is congruent to 7 modulo 19 (not 1) * -81 is congruent to 18 modulo 19 (not 1) There is only 1 integer in the set that is congruent to 1 modulo 19, which is -18. Therefore, the Legendre symbol (10/19) is equal to (-1)^1 = -1. Since the Legendre symbol is -1, we conclude that 10 is not a quadratic residue modulo 19. |
False |
| DeepSeek-Coder-V2 | To determine if 10 is a quadratic residue modulo 19 using Gauss's Lemma, we need to compute the Legendre symbol \(( \frac{10}{19} )\). According to Gauss's Lemma, this can be done by considering the least positive residues of the multiples of 10 modulo 19. First, let's list the least positive residues of the first 9 multiples of 10 modulo 19: 1. \(10 \times 1 \equiv 10 \mod 19\) 2. \(10 \times 2 \equiv 20 \equiv 1 \mod 19\) 3. \(10 \times 3 \equiv 30 \equiv 11 \mod 19\) 4. \(10 \times 4 \equiv 40 \equiv 2 \mod 19\) 5. \(10 \times 5 \equiv 50 \equiv 12 \mod 19\) 6. \(10 \times 6 \equiv 60 \equiv 3 \mod 19\) 7. \(10 \times 7 \equiv 70 \equiv 13 \mod 19\) 8. \(10 \times 8 \equiv 80 \equiv 4 \mod 19\) 9. \(10 \times 9 \equiv 90 \equiv 14 \mod 19\) Next, we count the number of these residues that are greater than \( \frac{19}{2} = 9.5 \): - 10 (greater than 9.5) - 11 (greater than 9.5) - 12 (greater than 9.5) - 13 (greater than 9.5) - 14 (greater than 9.5) There are 5 residues greater than 9.5. According to Gauss's Lemma, the Legendre symbol \(( \frac{10}{19} )\) is given by: \[ \left( \frac{10}{19} \right) = (-1)^k \] where \(k\) is the number of residues greater than 9.5. Here, \(k = 5\), so: \[ \left( \frac{10}{19} \right) = (-1)^5 = -1 \] Since the Legendre symbol \(( \frac{10}{19} ) = -1\), 10 is not a quadratic residue modulo 19. |
False |
| Qwen2-72B-Instruct | To determine if 10 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, modulo p. In this case, p = 19 and a = 10. So, we need to find the least positive residues of the first (19-1)/2 = 9 multiples of 10 modulo 19. 2. Count the number of least positive residues that are greater than p/2. If the number of such residues is even, then a is a quadratic residue modulo p. If the number of such residues is odd, then a is not a quadratic residue modulo p. Let's calculate the least positive residues of the first 9 multiples of 10 modulo 19: 1. 10 * 1 ≡ 10 (mod 19) 2. 10 * 2 ≡ 20 ≡ 1 (mod 19) 3. 10 * 3 ≡ 30 ≡ 11 (mod 19) 4. 10 * 4 ≡ 40 ≡ 2 (mod 19) 5. 10 * 5 ≡ 50 ≡ 12 (mod 19) 6. 10 * 6 ≡ 60 ≡ 3 (mod 19) 7. 10 * 7 ≡ 70 ≡ 14 (mod 19) 8. 10 * 8 ≡ 80 ≡ 4 (mod 19) 9. 10 * 9 ≡ 90 ≡ 16 (mod 19) Now, let's count the number of least positive residues that are greater than p/2 = 19/2 = 9.5: 1. 10 > 9.5 2. 11 > 9.5 3. 12 > 9.5 4. 14 > 9.5 5. 16 > 9.5 There are 5 least positive residues greater than 9.5, which is an odd number. Therefore, according to Gauss's Lemma, 10 is not a quadratic residue modulo 19. |
False |
| baichuan4 | To determine if 10 is a quadratic residue modulo 19 using Gauss's Lemma, we need to find an integer x such that x^2 ≡ 10 (mod 19). First, let's list the squares modulo 19: 0^2 ≡ 0 (mod 19) 1^2 ≡ 1 (mod 19) 2^2 ≡ 4 (mod 19) 3^2 ≡ 9 (mod 19) 4^2 ≡ 16 (mod 19) 5^2 ≡ 6 (mod 19) 6^2 ≡ 17 (mod 19) 7^2 ≡ 11 (mod 19) 8^2 ≡ 7 (mod 19) 9^2 ≡ 14 (mod 19) 10^2 ≡ 3 (mod 19) 11^2 ≡ 12 (mod 19) 12^2 ≡ 5 (mod 19) 13^2 ≡ 15 (mod 19) 14^2 ≡ 10 (mod 19) 15^2 ≡ 8 (mod 19) 16^2 ≡ 2 (mod 19) 17^2 ≡ 13 (mod 19) 18^2 ≡ 18 (mod 19) We see that 14^2 ≡ 10 (mod 19), so there exists an integer (14) such that its square is congruent to 10 modulo 19. Therefore, by Gauss's Lemma, 10 is a quadratic residue modulo 19. |
False |
| 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 = 10 and p = 19, we consider the set {10, 20, 30,...,(19-1)*10/2} = {10, 20, 30, 40, 50, 60, 70, 80, 90}. Of these, the numbers greater than 19/2 = 9.5 are 10, 20, 30, 40, 50, 60, 70, 80, 90. There are 9 such numbers. So m = 9. Then (-1)^9 = -1. So (10/19) = -1, which means 10 is not a quadratic residue modulo 19. |
False |
| 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 \( a = 10 \) is a quadratic residue modulo \( p = 19 \). We need to consider the multiples of 10 modulo 19, up to \( \frac{19-1}{2} = 9 \). We will look at the least positive residues of these multiples modulo 19: 1. \( 1 \cdot 10 \equiv 10 \mod 19 \) 2. \( 2 \cdot 10 \equiv 20 \equiv 1 \mod 19 \) 3. \( 3 \cdot 10 \equiv 30 \equiv 11 \mod 19 \) 4. \( 4 \cdot 10 \equiv 40 \equiv 2 \mod 19 \) 5. \( 5 \cdot 10 \equiv 50 \equiv 12 \mod 19 \) 6. \( 6 \cdot 10 \equiv 60 \equiv 3 \mod 19 \) 7. \( 7 \cdot 10 \equiv 70 \equiv 13 \mod 19 \) 8. \( 8 \cdot 10 \equiv 80 \equiv 4 \mod 19 \) 9. \( 9 \cdot 10 \equiv 90 \equiv 14 \mod 19 \) Now, we count the number of elements in the set \( \{10, 1, 11, 2, 12, 3, 13, 4, 14\} \) that are greater than \( \frac{19}{2} = 9.5 \). Since all numbers are less than 19, we only need to consider those greater than 9: - 10 is greater than 9 - 11 is greater than 9 - 12 is greater than 9 - 13 is greater than 9 - 14 is greater than 9 There are 5 numbers greater than 9 in the set, which is an odd number. Since the number of elements greater than \( \frac{p}{2} \) is odd, by Gauss's Lemma, 10 is not a quadratic residue modulo 19. |
False |
| kimi | A quadratic residue modulo \( n \) is an integer that is congruent to a perfect square modulo \( n \). Gauss's Lemma is a method used to determine whether a number is a quadratic residue modulo a prime number \( p \). The lemma states that a number \( a \) is a quadratic residue modulo \( p \) if and only if \( a \) has a square root modulo \( p \). This is equivalent to saying that \( a \) is a quadratic residue modulo \( p \) if and only if \( a \) has an inverse in the multiplicative group of integers modulo \( p \), which is denoted as \( (\mathbb{Z}/p\mathbb{Z})^* \). For a prime \( p \), the numbers \( 1, 2, \ldots, p-1 \) form a group under multiplication modulo \( p \), and each non-zero residue modulo \( p \) has a unique inverse in this group. However, the problem you've presented involves \( 10 \) modulo \( 19 \), which is not a prime number. Gauss's Lemma is specifically for prime moduli, so it cannot be directly applied to this problem. Instead, we can use the law of quadratic reciprocity or other methods to determine whether \( 10 \) is a quadratic residue modulo \( 19 \). The law of quadratic reciprocity is a more general result that can be used for any two distinct odd prime numbers. For the sake of simplicity, let's check if \( 10 \) is a quadratic residue modulo \( 19 \) by checking if \( 10^9 \equiv 1 \pmod{19} \), because \( 10 \) raised to the power of \( \frac{\phi(19)}{2} \) should be congruent to \( 1 \) modulo \( 19 \) if \( 10 \) is a quadratic residue modulo \( 19 \). Here, \( \phi(19) \) is Euler's totient function for \( 19 \), which is \( 18 \) since \( 19 \) is a prime number. Let's calculate \( 10^9 \mod 19 \): \[ 10^2 \equiv 100 \equiv 5 \pmod{19} \] \[ 10^4 \equiv 5^2 \equiv 25 \equiv 6 \pmod{19} \] \[ 10^8 \equiv 6^2 \equiv 36 \equiv 17 \pmod{19} \] \[ 10^9 \equiv 17 \cdot 10 \equiv 170 \equiv 18 \pmod{19} \] Since \( 10^9 \not\equiv 1 \pmod{19} \), \( 10 \) is not a quadratic residue modulo \( 19 \). |
False |
| wenxinyiyan | To determine if 10 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, let n be the least positive residue of $a \cdot 2^{-1}$ modulo p. Then, a is a quadratic residue modulo p if and only if the least positive residue of $n \cdot 2^{p-1}$ modulo p is even. In our case, $p = 19$ and $a = 10$. First, we find $2^{-1}$ modulo 19. This is the number that, when multiplied by 2, gives a result that is congruent to 1 modulo 19. It turns out that $2^{-1} \equiv 10 \mod 19$, because $2 \cdot 10 = 20 \equiv 1 \mod 19$. Next, we compute $n = 10 \cdot 10 \mod 19 = 100 \mod 19 = 5$. Now, we need to compute $5 \cdot 2^{18} \mod 19$. $5 \cdot 2^{18} \equiv 5 \cdot (2^9)^2 \equiv 5 \cdot (512)^2 \equiv 5 \cdot (18)^2 \equiv 5 \cdot 324 \equiv 5 \cdot 14 \equiv 70 \equiv 12 \mod 19$. The least positive residue of $5 \cdot 2^{18}$ modulo 19 is 12, which is even. Therefore, by Gauss's Lemma, 10 is a quadratic residue modulo 19. |
False |
| zhipu | To determine if 10 is a quadratic residue modulo 19 using Gauss's Lemma, we 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 \( a \) modulo \( p \) is even. ### Step 2: Set Up the Problem Here, \( p = 19 \) and \( a = 10 \). We need to check the integers from 1 to \( \frac{19-1}{2} = 9 \). ### Step 3: Find the Congruences We need to find the integers \( k \) such that \( 1 \leq k \leq 9 \) and \( 10k \equiv m \pmod{19} \) for some \( m \) in the range \( 1 \leq m \leq 9 \). Let's calculate \( 10k \mod 19 \) for \( k = 1, 2, \ldots, 9 \): - \( 10 \cdot 1 = 10 \equiv 10 \pmod{19} \) - \( 10 \cdot 2 = 20 \equiv 1 \pmod{19} \) - \( 10 \cdot 3 = 30 \equiv 11 \pmod{19} \) - \( 10 \cdot 4 = 40 \equiv 2 \pmod{19} \) - \( 10 \cdot 5 = 50 \equiv 12 \pmod{19} \) - \( 10 \cdot 6 = 60 \equiv 3 \pmod{19} \) - \( 10 \cdot 7 = 70 \equiv 13 \pmod{19} \) - \( 10 \cdot 8 = 80 \equiv 4 \pmod{19} \) - \( 10 \cdot 9 = 90 \equiv 14 \pmod{19} \) ### Step 4: Count the Congruences Now, we count how many of these values are greater than \( \frac{19}{2} = 9.5 \): - \( 10, 11, 12, 13, 14 \) There are 5 values greater than 9.5. ### Step 5: Apply Gauss's Lemma According to Gauss's Lemma, if the number of these values is even, then \( a \) is a quadratic residue modulo \( p \). If it is odd, then \( a \) is not a quadratic residue modulo \( p \). Here, the number of values greater than 9.5 is 5, which is odd. ### Conclusion Since 5 is odd, 10 is not a quadratic residue modulo 19. Thus, **10 is not a quadratic residue modulo 19**. |
False |
| gpto1 | To determine whether 10 is a quadratic residue modulo 19 using **Gauss's Lemma**, we'll follow these steps: **Gauss's Lemma Statement:** For an odd prime \( p \) and an integer \( a \) not divisible by \( p \), let \( n \) be the least positive residue of \( a \) modulo \( p \). Consider the set: \[ S = \{ n, 2n, 3n, \dotsc, \left( \frac{p-1}{2} \right) n \} \] Reduce each element of \( S \) modulo \( p \) to get their least positive residues. Let \( m \) be the number of these residues that are greater than \( \frac{p}{2} \). Then: \[ \left( \frac{a}{p} \right) = (-1)^m \] where \( \left( \frac{a}{p} \right) \) is the **Legendre symbol**. If \( \left( \frac{a}{p} \right) = 1 \), then \( a \) is a quadratic residue modulo \( p \); if \( \left( \frac{a}{p} \right) = -1 \), then \( a \) is a quadratic non-residue modulo \( p \). **Step-by-Step Solution:** 1. **Identify Parameters:** - \( p = 19 \) (an odd prime) - \( a = 10 \) - \( n = 10 \) (since \( a \mod p = 10 \)) 2. **Compute \( \frac{p - 1}{2} \):** \[ \frac{19 - 1}{2} = 9 \] So we need to consider multiples of \( n \) from \( 1 \) to \( 9 \). 3. **Construct the Set \( S \):** \[ S = \{ n, 2n, 3n, \dotsc, 9n \} = \{10, 20, 30, \dotsc, 90\} \] 4. **Compute Least Positive Residues Modulo \( p \):** Reduce each element in \( S \) modulo \( 19 \): \[ \begin{align*} s_1 &= 10 \mod 19 = 10 \\ s_2 &= 20 \mod 19 = 1 \\ s_3 &= 30 \mod 19 = 11 \\ s_4 &= 40 \mod 19 = 2 \\ s_5 &= 50 \mod 19 = 12 \\ s_6 &= 60 \mod 19 = 3 \\ s_7 &= 70 \mod 19 = 13 \\ s_8 &= 80 \mod 19 = 4 \\ s_9 &= 90 \mod 19 = 14 \\ \end{align*} \] 5. **Determine Which Residues Are Greater Than \( \frac{p}{2} \):** Calculate \( \frac{p}{2} \): \[ \frac{19}{2} = 9.5 \] List the residues and mark those greater than 9.5: - \( s_1 = 10 \) (> 9.5) - \( s_2 = 1 \) (< 9.5) - \( s_3 = 11 \) (> 9.5) - \( s_4 = 2 \) (< 9.5) - \( s_5 = 12 \) (> 9.5) - \( s_6 = 3 \) (< 9.5) - \( s_7 = 13 \) (> 9.5) - \( s_8 = 4 \) (< 9.5) - \( s_9 = 14 \) (> 9.5) Residues greater than \( \frac{p}{2} \) are \( s_1, s_3, s_5, s_7, s_9 \). 6. **Count \( m \):** Number of residues greater than \( 9.5 \): \[ m = 5 \] 7. **Compute the Legendre Symbol Using Gauss's Lemma:** \[ \left( \frac{10}{19} \right) = (-1)^m = (-1)^5 = -1 \] 8. **Conclusion:** Since \( \left( \frac{10}{19} \right) = -1 \), the number \( 10 \) is a **quadratic non-residue** modulo \( 19 \). **Answer:** **No, 10 is not a quadratic residue modulo 19; using Gauss's Lemma, we find that (10 | 19) = –1.** |
False |
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