Question

1.
Let [x] denote the integer part of x.
( eg. [2.56788] = 2 )
Prove that [(2 + √3)^n] is an odd number, where n is a non-negative integer.

2.
Let {x} denote the decimal part of x. ( eg. {2.44598} = 0.44598 )
If for natural number n , R = (2 + √3)^n and f = {R} ,
Prove that R(1 - f) = 1

Answer 1

Do you need a hit?

Answer 2

not yet

Answer 3

lol ... we don't need 'hit'

Answer 4

haha hint

Answer 5

where do you get your problems ... most of them are insanely difficult

Answer 6

internet

Answer 7

and Hong Kong Advanced Level Examination (Pure Math)

Answer 8

These two questions are not :)

Answer 9

Okay. I got it.
Should I post my solution or do guys want to try it too?

Answer 10

can't wait

Answer 11

Haha. Okay.
Brb.

Answer 12

so sid bring ur answer

Answer 13

For convinience I type 3^1/2 as k.
So, Let (2 + k) ^ n be some I + f1 ---- Where I is the integral part and f the fractional.

And Let ( 2 - k)^n be some f2 ( It has to be fractional)
Adding the two we get,
I1 + + f1 + f2 = ( Something which is integral, As on adding the irrational parts cancel out to give only an integer.)
Therefore, f1 + f2 + I1is an integer. => f1 + f2 is an integer.
Now 0<f1<1 and f2 also has the same range. Therefore 0<f1+f2<2.
For it to be an integer f1 + f2 = 1.
Therefor I1 + 1 = even. ( Its Twice of something)
Therefore I1 = Greatest integer of given expression is odd.