# Find the Set W

• Aug 30th 2012, 07:56 AM
brucewayne
Find 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}.
• Aug 31st 2012, 04:33 AM
brucewayne
Re: Find the Set W
such that (x^3-3x+2)/(2x+1)
• Aug 31st 2012, 05:56 AM
emakarov
Re: 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...
• Aug 31st 2012, 06:27 AM
Deveno
Re: Find the Set W
suppose that (x3 - 3x + 2)/(2x + 1) = k, where k is an integer.

then (8x3 - 24x + 16)/(2x + 1) = 8k. this is also an integer.

now 8x3 - 24x + 16 = (2x + 1)(4x2 - 2x - 11) + 27

so 8k = (8x3 - 24x + 16)/(2x + 1) = 4x2 - 2x - 11 + 27/(2x - 1).

since for ALL integers x, 4x2 - 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.
• Aug 31st 2012, 04:26 PM
brucewayne
Re: 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
• Aug 31st 2012, 05:48 PM
Deveno
Re: 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.