siuhochung 寫到:[u^k] congruent to k-1 (mod 2).
can i do it in this way?
Let u be a root of x^2-ax-b=0
let a,b E Z, and v be the another root of this equation.
Let S(k)=u^k+v^k
It represents a recurrence relation, therefore S(k) is integer for k E N.
When a=0(mod 2), u=v(mod 2), u^k+v^k=0(mod 2), S(k)=0(mod 2)
Take a,b such that -1<v<0, then
S(k)<u^k<S(k+1), if k is odd.
S(k-1)<u^k<S(k), if k is even.
So [u^k]=s(k) is even, if k is odd.
=S(k-1) is odd, if k is even.
v= [a-sqrt(a^2-4b)]/2 ,
-a-sqrt(a^2-4b)>-2
a+sqrt(a^2-4b)<2
a^2-4b<a^2-4a+4
-b<-a+1
a<b+1
a=0(mod 2)
Say a=-2, b=-1
Then
x^2-2x-1=0
u = (2 + sqrt(8)/2
u = 1+sqrt2
Can I do it in this way?
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Take a,b such that -1<v<0, then
S(k)<u^k<S(k)+1, if k is odd.
S(k)-1<u^k<S(k), if k is even.
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-1<v= [a-sqrt(a^2+4b)]/2 <0,
a+2>sqrt(a^2+4b) >a
if a=2k>=0,
aa+4a+4> aa+4b> aa
2k>= b> =1
x^2-2kx-b=0
x=k+sqrt(kk+b)
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when k =1: b=1 or 2, x=1+sqrt(2) or 1+sqrt(3)
when k=2: b=1,2,3, or 4, x=2+sqrt(5) ,2+sqrt(6), 2+sqrt(7) or 2+sqrt(8)
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