Question

Please give the complete steps, thank you

3. If X1, X, are the zeros of the polynomial x2 - 6x + 1, then for every nonnegative integer n, x + x is an integer and not divisible by 5. Clue 3. We have sı = x1 + x2 = 6, X;X2 = 1. Let S = x+x. In the section on symmetric polynomials, we established that s = 65-1 - S-2. Starting from so = 2, $y = 6, this recurrence gives only integral values. Consider the s, modulo 5. The recurrence becomes Sn = $m-1 - S7-2. We get $0 =2, $1=1, $2 = 4, $3 = 3, $4 = 4, 5 1 , = 2, $y = 1,.... After six steps, the pair (2, 1) recurs and the sequence is periodic without any zero (multiple of 5). Remark. The characteristicequation of the sequences, 65,-1-S, -is.x2 - 6x +1 = 0, and the general solution is sy =(3+ 8) + (3 - 78)*.

For number 3, the question is if x_1, x_2 are the zeros of the polynomial x^2-6x+1

solve it with the given clue, i need the clear steps to solve, thank you.

Answer 1

Given x, and x₂ ore the zones of 3 - 6xllzo → sum of 3tava * ****= 6 =S; (ray) — and product of zoos en ,42=1. = So we So = 1+1=2,5,6 established that Sn = 650-1 - Snea. - Sn = Song -Sned (med 5) Now Now (1-2)} = (0,47,– 47, - 36 - 4 = 32 455 ☺ → X-X2 = ਨਾਮ 0 and . - . 34 212 a = 3-25 i Sn = (3 +257+ (3 - 256)" 6 which is . si - . 3+2/2+3-2/2 = for hal sa 3+2/2+3-2/2 = not divisible by 5. but si in integer. sn always an integer & n. so