Consider the equation (E) : 11x â 7y = 5 with x, y integers. (a) Give a solution of 11x â 7y = 1. ... (a) Suppose that (x, y) is a solution of (F). Show that x2 = 2y2(5).
Exercises BAC Exercise 1 : 1. Show that, for all integer n greater than or equal to 2, HCF (3n3 − 11n, n + 3) = HCF (48, n + 3). 2. Using the previous question, deduce the possible values of HCF (3n3 − 11n, n + 3). 3n3 − 11n is an integer. n+3 3n3 − 11n is in its simplest form. 4. Give a condition on the integer n such that the fraction n+3 3. Determine the set of integers n such that the fraction
Exercise 2 : 1. Consider the equation (E) : 11x − 7y = 5 with x, y integers. (a) Give a solution of 11x − 7y = 1. Give the name of the theorem we can use to assert that such a solution exists. (b) Deduce a solution of 11x − 7y = 5. (c) Solve the equation (E). (d) In a orthonormal coordinate system (O,~i, ~j) of the plane, consider the line D of equation 11x − 7y − 5 = 0 and C the set of points M (x, y) such that 0 < x < 50 and 0 < y < 50. How many points are in the intersection set of D and C ? 2. Consider the equation (F ) : 11x2 − 7y 2 = 5 with integers x, y. (a) Suppose that (x, y) is a solution of (F ). Show that x2 = 2y 2 (5). (b) Give the table of congruence modulo 5 of x2 and 2x2 . (c) Deduce that if (x, y) is a solution of (F ) then 5 divides x and y. 3. Suppose that 5 divides x and y. Is the pair (x, y) a solution of (F ) ? Explain. 4. Solve (F ). Exercise 3 : Let (E) be the Diophantine equation 16x − 3y = 4. 1. Give one possible solution of (E). 2. Solve (E). Exercise 4 : Consider the equation (E) : 109x − 226y = 1 with integers x, y. 1. Calculate HCF (109, 226) and deduce a possible solution of (E). 2. Solve (E). 3. Show that there is a unique pair of integers (d, e) such that 109d = 1+226e and d < 226. Give the values of d and e. Exercise 5 : This is a TRUE/FALSE. Explain your answers. 1. The equation 24x + 34y = 2 has solutions {(34k − 7, 5 − 24k) s.t. k ∈ Z} 2. If we can find integers u and v (both greater than or equal to 1) such that au + bv = 2 then we can state that HCF (a, b) = 2. 11x 1 3. The line of equation y = − has an infinity of points with integers coordinates. 15 5
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