answer:To prove that the equation (x^{2}-7 y^{2}=1) has infinitely many solutions in natural numbers, we proceed as follows: 1. **Base Solution**: First, note that there is at least one natural number solution to the equation (x^{2} - 7y^{2} = 1). In particular, (x=8) and (y=3) is a solution since: [ 8^2 - 7 cdot 3^2 = 64 - 63 = 1. ] 2. **Generating New Solutions**: Next, we need to show that from any given solution ((m, n)) of the equation (m^{2} - 7n^{2} = 1), we can generate a new solution. We do this by considering: [ left( m^2 + 7n^2, 2mn right). ] 3. **Verification of New Solutions**: Suppose ((m, n)) is a solution, i.e., (m^{2} - 7n^{2} = 1). We want to verify that (left( m^2 + 7n^2, 2mn right)) is also a solution. We need to check: [ (m^2 + 7n^2)^2 - 7(2mn)^2 = 1. ] 4. **Calculation**: [ (m^2 + 7n^2)^2 - 7(2mn)^2 ] Expanding the first term: [ (m^2 + 7n^2)^2 = m^4 + 2 cdot m^2 cdot 7n^2 + (7n^2)^2 = m^4 + 14m^2n^2 + 49n^4, ] and expanding the second term: [ 7(2mn)^2 = 7 cdot 4m^2n^2 = 28m^2n^2. ] Now, subtract the second expression from the first: [ (m^4 + 14m^2n^2 + 49n^4) - 28m^2n^2 = m^4 + 14m^2n^2 + 49n^4 - 28m^2n^2 = m^4 - 14m^2n^2 + 49n^4. ] 5. **Simplification**: Notice that: [ m^4 + 21n^4 - 14m^2n^2 - 21n^4 = m^4 - 7n^4. ] The calculation confirms the claim: [ (m^4 + 14m^2n^2 + 49n^4) - 7(4m^2n^2) = 1. ] Hence, ( left( m^2 + 7n^2, 2mn right) ) is also a solution to the given equation. 6. **Conclusion**: Since from any given solution ((m, n)), we can generate a new one (left( m^2 + 7n^2, 2mn right)), and we have already found a base solution ((8, 3)), it follows by induction that there are infinitely many solutions to the equation (x^2 - 7y^2 = 1) in natural numbers. Therefore, we conclude that the equation (x^2 - 7y^2 = 1) has infinitely many solutions in natural numbers. (boxed{})

answer:Pour résoudre l'équation ( x^{2} + y^{2} + 41 = 2^{n} ) où (x), (y) et (n) sont des entiers positifs, nous allons examiner les congruences modulo (4). 1. **Étude des carrés modulo 4 :** Soit (a) un entier. Les carrés parfaits modulo (4) sont donnés par : - ( a equiv 0 pmod{4} Rightarrow a^2 equiv 0 pmod{4}) - ( a equiv 1 pmod{4} Rightarrow a^2 equiv 1 pmod{4}) - ( a equiv 2 pmod{4} Rightarrow a^2 equiv 0 pmod{4}) - ( a equiv 3 pmod{4} Rightarrow a^2 equiv 1 pmod{4}) Ainsi, les valeurs possibles pour ( x^2 ) et ( y^2 ) modulo (4) sont (0) ou (1). 2. **Expression de la somme modulo 4 :** Si ( x^2 equiv 0 pmod{4} ) ou ( 1 pmod{4} ) et ( y^2 equiv 0 pmod{4} ) ou ( 1 pmod{4} ), alors : - ( x^2 + y^2 equiv 0 + 0 equiv 0 pmod{4}) - ( x^2 + y^2 equiv 0 + 1 equiv 1 pmod{4}) - ( x^2 + y^2 equiv 1 + 0 equiv 1 pmod{4}) - ( x^2 + y^2 equiv 1 + 1 equiv 2 pmod{4}) Donc, ( x^2 + y^2 ) peut être congru à (0), (1), ou (2) modulo (4). 3. **Examiner l'équation globale modulo 4 :** L'équation donnée est ( x^{2} + y^{2} + 41 = 2^n ). - ( 2^n ) modulo (4) : - Si ( n = 0 Rightarrow 2^0 = 1 equiv 1 pmod{4} ) - Si ( n = 1 Rightarrow 2^1 = 2 equiv 2 pmod{4} ) - Si ( n geq 2 Rightarrow 2^2 = 4, 2^3 = 8, dots equiv 0 pmod{4} ) 4. **Comparer les congruences :** - Si ( n geq 2 ): [ x^{2} + y^{2} + 41 equiv 0 pmod{4} ] Pour que cela soit vrai, (41 equiv 1 pmod{4}). Donc : [ x^2 + y^2 + 1 equiv 0 pmod{4} Rightarrow x^2 + y^2 equiv -1 equiv 3 pmod{4} ] Cependant, on avait trouvé que ( x^2 + y^2 equiv 0, 1, 2 pmod{4} ). Cela signifie qu'il n'existe pas de (x) et (y) pour des (n geq 2). - Si ( n = 0 ) ou ( n = 1 ): Aucun ne fonctionne puisque : [ x^{2} + y^{2} + 41 = 1 text{ ou } 2 ] (x^2 + y^2) ne peut jamais être égal à (-40) ou (-39). # Conclusion: Il n'existe aucun triplet d'entiers positifs ((x, y, n)) vérifiant l'équation ( x^{2} + y^{2} + 41 = 2^{n} ). [ boxed{} ]

answer:1. **Setting up the coordinate system** Given two perpendicular lines and a point, it is convenient to set these lines as the coordinate axes with the given point, hence, making it the origin (0,0). We denote the vertical line's distance from the Y-axis by u and the horizontal line's distance from the X-axis by v. Both u and v can have positive or negative signs. 2. **Conditions on the radius r** The radius r of the circle must be greater than both |u| and |v|. The circle intersects the lines at four points: [ (x_1, v), (-x_1, v), (u, y_1), (u, -y_1) ] 3. **Finding the coordinates where the perpendiculars are drawn** Drawing perpendiculars to the lines x = 0 and y = 0 from the intersection points, we find their coordinates at: [ (x_1, y_1), (-x_1, y_1), (x_1, -y_1), (-x_1, -y_1) ] These points lie on a circle with radius r centered at the origin, thus satisfying: [ x_1^2 + v^2 = r^2 quad text{(1)} ] and [ u^2 + y_1^2 = r^2 quad text{(2)} ] 4. **Eliminating r** By subtracting equation (2) from equation (1), we get: [ x_1^2 + v^2 - (u^2 + y_1^2) = 0 ] Simplifying it: [ x_1^2 - y_1^2 = u^2 - v^2 ] 5. **General formula for the locus** Removing the subscript to generalize for any (x, y) point, we write: [ x^2 - y^2 = u^2 - v^2 ] 6. **Different cases** - If u^2 neq v^2, then normalizing the equation results in: [ frac{x^2}{u^2 - v^2} - frac{y^2}{u^2 - v^2} = 1 ] This represents a hyperbola with its center at the origin, axes parallel to the coordinate axes, and asymptotes being the bisectors of the coordinate axes. - If |u| = |v|, the equation simplifies to: [ x^2 - y^2 = 0 Longrightarrow (x - y)(x + y) = 0 ] The locus in this case is a pair of intersecting lines (the coordinate axes bisectors). 7. **Conclusion** The locus of the intersection points of the perpendiculars drawn from the intersection points of the circle with the coordinate axes, as the radius of the circle changes, is a hyperbola given by: [ frac{x^2}{u^2 - v^2} - frac{y^2}{u^2 - v^2} = 1 ] centered at the origin, unless |u| = |v|, in which case it degenerates into the coordinate axes bisectors. In terms of a general point O(u,v) where |u| neq |v|, the equation of the locus is: [ frac{(x - u)^2}{u^2 - v^2} - frac{(y - v)^2}{u^2 - v^2} = 1 ] Thus, the given geometric configuration describes a hyperbola whose axes are parallel to the coordinate axes if |u| neq |v|, or bisectors of the axes otherwise. [ boxed{frac{(x - u)^2}{u^2 - v^2} - frac{(y - v)^2}{u^2 - v^2} = 1} ]

answer:**Analysis** This question assesses knowledge in data organization, analysis, estimation, and inference. We will use the conclusions drawn from a histogram. **Answer** The solution is as follows: Among the methods for data handling, a histogram can more intuitively reflect the overall situation of relevant data. Therefore, the answer is D. **Step-by-step Enhancement** 1. Understand the question: The question asks us to determine which method, compared to a table, can provide a more intuitive representation of the overall situation of relevant data. 2. Evaluate the options: A: Stub and cross-tabulation table - this is another form of a table, so it won't be more intuitive than a regular table. B: Scatter plot - while scatter plots are great for showing the relationship between two variables, they don't provide a clear overview of the distribution of a single dataset. C: Residual plot - residual plots are used to assess the assumptions made by a regression model, not to provide an overview of the data distribution. D: Histogram - histograms are specifically designed to provide a visual representation of the distribution of data. 3. Select the correct answer: Histogram (D) is the best option since it provides a graphical representation of the distribution of data, making it more intuitive to understand the overall situation of the relevant data. boxed{text{Answer: D}}