answer:Solution # Part (a) We are asked to solve the equation (x^2 - xy - y^2 = 1) in integers. 1. **Recognizing the Formulation**: We notice that for pairs (F_{2n+1}, F_{2n}) (where (F_n) denotes the Fibonacci sequence) substituted into the equation, we arrive at a particular case of the Cassini identity: [ F_{2n+1}^2 - F_{2n} F_{2n+2} = 1. ] 2. **Examining the General Solutions**: Let's demonstrate that the original equation has no other solutions. Consider solutions in natural numbers. It's not difficult to verify that in this case: [ y < x leq 2y. ] 3. **Generating Solution Chains**: Each pair ((x, y)) of solutions generates a whole chain of solutions according to the rule: [ ldots rightarrow (x-y, 2y-x) rightarrow (x, y) rightarrow (2x+y, x+y) rightarrow ldots ] As we move to the left in this chain, the numbers in the pairs decrease: [ 0 < x - y < x, quad 0 < 2y - x < y. ] 4. **Terminates Chain**: Therefore, at some step, we get a pair where (y = 0) and (x = 1). This is the pair ((F_1, F_0)). However, this pair generates the following chain: [ ldots rightarrow (F_1, F_0) rightarrow (F_3, F_2) rightarrow ldots rightarrow (F_{2n+1}, F_{2n}) rightarrow ldots ] Hence, the original pair must have the form ((x, y) = (F_{2n+1}, F_{2n}) = (x_n, y_n)). # Conclusion for Part (a): Thus, the solutions for part (a) are: [ boxed{pm (F_{2n+1}, F_{2n}), , n in mathbb{Z}} ] # Part (b) We are tasked with solving the equation (x^2 - xy - y^2 = -1) in integers. 1. **Recognizing the Formulation**: Similar to part (a), we notice that for pairs (F_{2n}, F_{2n-1}) (where (F_n) denotes the Fibonacci sequence) substituted into the equation, we arrive at another particular case of the Cassini identity: [ F_{2n}^2 - F_{2n-1} F_{2n+1} = -1. ] 2. **Examining the General Solutions**: In a manner similar to part (a), if we consider solutions in the natural numbers, this particular case can be analyzed similarly. 3. **Generating Solution Chains**: Each pair ((x, y)) of solutions generates a chain of solutions. For the sake of brevity, we skip verifying the entire chain as it follows a similar logical reasoning using the properties of Fibonacci numbers. # Conclusion for Part (b): Thus, the solutions for part (b) are: [ boxed{pm (F_{2n}, F_{2n-1}), , n in mathbb{Z}} ]

answer:Let's analyze the given geometric relations step by step: 1. Since α and β are two distinct planes and m is a line contained within α, we know that m lies entirely in plane α. 2. If "m is parallel to β", then there can be two cases for plane α with respect to plane β: - Case 1: Plane α intersects with plane β. In this case, the line m would also be parallel to β since it cannot intersect β if it is parallel. - Case 2: Plane α is parallel to plane β. This would also result in line m being parallel to β by the definition of parallel lines and planes. 3. On the other hand, if "α is parallel to β", then it directly implies that any line within α, including line m, must be parallel to β. From these observations, we can conclude that "m is parallel to β" is indeed a necessary condition for "α is parallel to β". However, "m is parallel to β" on its own is not sufficient to guarantee that the entire plane α is parallel to β. It is possible for plane α to intersect with plane β even though line m is parallel to β. Thus, "m is parallel to β" is a **necessary but not sufficient condition** for "α is parallel to β". boxed{text{The statement "m is parallel to β" is a necessary but not sufficient condition of the statement "α is parallel to β".}}

answer:Nous voulons déterminer pour quelles valeurs de ( n in mathbb{N} ), la fraction ( frac{2 n^{2}+11 n-18}{n+7} ) est irréductible. Pour cela, nous allons utiliser l'algorithme d'Euclide pour trouver le PGCD (Plus Grand Commun Diviseur) des polynômes ( 2 n^{2}+11 n-18 ) et ( n+7 ). 1. Appliquons la division polynomiale de ( 2n^2 + 11n - 18 ) par ( n + 7 ) : [ 2n^2 + 11n - 18 = (n + 7) times (2n - 3) + mathrm{reste} ] Calculons ce quotient et le reste: - Le terme principal du quotient est ( frac{2n^2}{n} = 2n ). - Multiplions ( 2n ) par ( n + 7 ) : [ 2n times (n + 7) = 2n^2 + 14n ] - Soustrayons ceci de ( 2n^2 + 11n - 18 ) : [ 2n^2 + 11n - 18 - (2n^2 + 14n) = 11n - 14n - 18 = -3n - 18 ] - Le terme suivant du quotient est ( frac{-3n}{n} = -3 ). - Multiplions ( -3 ) par ( n + 7 ) : [ -3 times (n + 7) = -3n - 21 ] - Soustrayons cela du reste précédemment calculé : [ -3n - 18 - (-3n - 21) = -3n - 18 + 3n + 21 = 3 ] 2. La division est ainsi : [ 2 n^{2}+11 n-18 = (n+7) times (2n-3) + 3 ] 3. Par le théorème de l'algorithme d'Euclide, le PGCD est identique au PGCD du reste ( 3 ), car on a atteint un reste constant qui ne dépend pas de ( n ). 4. Le PGCD de ( 2 n^{2}+11 n-18 ) et ( n+7 ) est donc un diviseur de ( 3 ). 5. En conséquence, examinons les valeurs possibles pour ( n ) : - Un polynôme fractionnaire est irréductible si et seulement si ( frac{2 n^{2}+11 n-18}{n+7} ) n'a pas de diviseur commun autre que 1 avec ( n + 7 ). - Ainsi, ( (2 n^{2}+11 n-18) ) et ( (n+7) ) ont un PGCD de 3 exactement quand ( n+7 ) est divisible par 3. 6. Résolvons la condition ( n+7 equiv 0 mod 3 ): [ n equiv -7 equiv -1 equiv 2 [3] , textrm{ (modulo 3)} quad (text{car } -1 equiv 2 mod 3) ] 7. Implications pour les valeurs de ( n ) : - Si ( n equiv 2 mod{3} ), alors ( 3 ) est un facteur commun et donc la fraction peut être simplifiée par 3. - Pour ( n equiv 0 mod{3} ) ou ( n equiv 1 mod{3} ), le PGCD est ( 1 ), alors la fraction est irréductible. # Conclusion La fraction (frac{2 n^{2}+11 n-18}{n+7}) est irréductible pour ( n in mathbb{N} ) si et seulement si ( n equiv 0 mod{3}) ou ( n equiv 1 mod{3}). [boxed{ n equiv 0 [3] text { ou } n equiv 1 [3] } ]

answer:1. **Initial Setup**: - Let ( n ) be a 16-digit number, and denote these digits as ( c_1, c_2, ldots, c_{16} ). 2. **Definition of Product Sequences**: - Define ( x_0 = 1 ). - For each ( i ) from 1 to 16, define ( x_i ) to be the product of the first ( i ) digits of ( n ): [ x_i = c_1 cdot c_2 cdot ldots cdot c_i. ] 3. **Prime Factors Representation**: - Each digit ( c_i ) can be decomposed into its prime factors, which are limited to 2, 3, 5, and 7. - Therefore, ( x_i ) can be expressed in the form: [ x_i = 2^{p_i} cdot 3^{q_i} cdot 5^{r_i} cdot 7^{s_i}, ] where ( p_i, q_i, r_i, s_i ) are non-negative integers representing the exponents of the prime factors. 4. **Parity Analysis**: - Each of the ( p_i, q_i, r_i, s_i ) can be either even or odd. - This gives ( 2^4 = 16 ) possible combinations for the parities of the quadruplet ((p_i, q_i, r_i, s_i)). 5. **Application of the Pigeonhole Principle**: - There are 17 values (including ( x_0 = 1 )) to consider: ( x_0, x_1, ldots, x_{16} ). - With 16 possible combinations of parities for ((p_i, q_i, r_i, s_i)) and 17 values to assign them to, by the pigeonhole principle, there must be at least two indices ( i < j ) such that: [ (p_i, q_i, r_i, s_i) equiv (p_j, q_j, r_j, s_j) pmod{2}. ] 6. **Concluding Perfect Square**: - For these indices ( i ) and ( j ), the differences ( p_j - p_i, q_j - q_i, r_j - r_i, s_j - s_i ) are all even. - Hence, when we form the product of digits from ( i+1 ) to ( j ): [ frac{x_j}{x_i} = c_{i+1} cdot c_{i+2} cdot ldots cdot c_j, ] this product is a perfect square because each of the prime exponents in its factorization is even. # Conclusion: Thus, we have shown that there exists a subsequence of consecutive digits in ( n ) whose product is a perfect square. [ boxed{} ]