中四附加數(15pts)

1. Let k and 2k be the two roots of the equation.
sum of roots = k+2k = -b/a
3k = -b/a
k = -b/3a
Product of roots = 2k^2 = c/a
2(-b/3a)^2 = c/a
2b^2/9a^2 = c/a
2b^2 = 9ac

2. Let k1 and k2 be the roots of bx^2+cx+a=0
k1+k2 = -c/b and k1k2 = a/b
If roots of ax^2+bx+c=0 is three times the roots of bx^2+cx+a=0, i.e.
-b/a = -3c/b and c/a = 9a/b
b^2 = 3ac and 9a^2 = bc

3. a+b = 1, ab = -1
a-b = root((a+b)^2 -4ab) = sqrt 5
f(1) = 1/sqrt5 (a-b) = 1/sqrt5* sqrt5 = 1
f(2) = 1/sqrt5 (a^2-b^2) = 1/sqrt5 *(a+b)(a-b) = 1

(b) f(n)+f(n+1) = 1/sqrt5 [a^n-b^n+a^(n+1)-b^(n+1)]
=1/sqrt5 [a^n+a^(n+1)-b^n-b^(n+1)]
=1/sqrt5 [a^(n+2)-b^(n+2)] ***Note that a and b are the roots of x^2-x-1=0, i.e. a^2-a-1=0 and b^2-b-1=0 hence a^(n+2)-a^(n+1)-a^n = 0 and b^(n+2)-b^(n+1)-b^n = 0
Therefore, a^n+2 = a^n+a^(n+1) and b^n+2 = b^n+b^(n+1) ***
=f(n+2)

(c) By (a), f(1) = f(2) = 1, which is integer.
The statement is true for n = 1 and 2.
Assume the statement is true for n = k and n = k+1, i.e. f(k) and f(k+1) are integers.
f(k+2) = f(k)+f(k+1) must be an integer.
The statement is true for n = k+2.
By M.I. the statement is true for all positive integers n.