I need a few hints. Find the set W= {x belongs to Z (integers) such that

http://www2.wolframalpha.com/Calcula...&s=6&w=79&h=39 belongs to the integers}.

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- August 30th 2012, 07:56 AMbrucewayneFind the Set W
I need a few hints. Find the set W= {x belongs to Z (integers) such that

http://www2.wolframalpha.com/Calcula...&s=6&w=79&h=39 belongs to the integers}. - August 31st 2012, 04:33 AMbrucewayneRe: Find the Set W
such that (x^3-3x+2)/(2x+1)

- August 31st 2012, 05:56 AMemakarovRe: Find the Set W
By testing numbers whose absolute value is < 10^6, it seems that W = {-14, -5, -2, -1, 0, 1, 4, 13}, but I am not sure how to prove it...

- August 31st 2012, 06:27 AMDevenoRe: Find the Set W
suppose that (x

^{3}- 3x + 2)/(2x + 1) = k, where k is an integer.

then (8x^{3}- 24x + 16)/(2x + 1) = 8k. this is also an integer.

now 8x^{3}- 24x + 16 = (2x + 1)(4x^{2}- 2x - 11) + 27

so 8k = (8x^{3}- 24x + 16)/(2x + 1) = 4x^{2}- 2x - 11 + 27/(2x - 1).

since for ALL integers x, 4x^{2}- 2x - 11 is an integer, if 8k is to be an integer, we must have 2x - 1 is a divisor of 27, so:

2x - 1 = ±1, 3, 9 or 27.

this leads to W = {-14,-5,-2,-1,0,1,4,13} as emakarov conjectured. - August 31st 2012, 04:26 PMbrucewayneRe: Find the Set W
This is where I always get stumped. I would have never guessed to multiply by 8. I just started really learning mathematics about two years ago, and I have a difficult time introducing numbers that help in solving the equation. Why 8 and not 3, 4 or some other number? Just curious. Thanks

- August 31st 2012, 05:48 PMDevenoRe: Find the Set W
well first, i just did ordinary polynomial division over the rationals. then i took the least common denominator of all the coefficients i got to turn everything into integers. that least common denominator just happened to be 8.