Prove that there is no integer n such that
$\displaystyle \frac{1}{n^{3}-1}+ \frac{1}{n^{3}+1} = \frac{1}{n} $
Can I just say, when n = 1 it it does not work for '1' than it must not work for all other integers?
Clear fractions, then set it up as a polynomial for which you're looking for the roots. If it had an integer solution a, then even without solving that polynomial, you could could restrict what a has to be to just a few choices. None of those choices are solutions, and therefore, there are no integer solutions to that equation.
Ex: Does x^2 - 7x +3 = 0 have an integer solution? You can answer that w/o solving that quadratic. If it had an integer solution a, then it would've factored with (x-a) being a factor (and the other also being an integer - this is a theorem). Thus a would divide 3.
So if this has an integer solution, it must be in the set {-1, 1, -3, 3}. But none of those is a solution, which you can see by just checking them. Therefore, that polynomial doesn't have an integer root.
Thank you for that,
I have got rid of the brackets and got $\displaystyle 2n^{6} - 2n^{3} -1 = 0 $
But I dont see how I can tell wheather this polynomial has an integer solution or not?
So if this polynomial had an integer solution 'a', (x-a), than a would divide -1, ?
I believe it's $\displaystyle n^6 - 2n^4 - 1 = 0$
The main rule is:
If the leading coefficent is 1 (the case above, it's $\displaystyle 1n^6$),
and all the coefficients are integers,
then any integer solution will divide the constant term.
So in this case, the only possible integer solutions are {-1, 1}, and neither of those are solutions, so it has no integer solutions.