how come it only works with n=3? there must be some sort of reasoning behind it? because say if i had not known that n=3 is a solution (by inputting values into excel) then how would i have come about the answer?
also, in the proof for n>3 this has been used:
CASE I: n is even,
(n^{2}+\frac{n}{2})^{2}<\frac{n^{5}-1}{n-1}< (n^{2}+\frac{n}{2}+1)^{2}
CASE II: n is odd,
(n^{2}+\frac{n-1}{2})^{2}<\frac{n^{5}-1}{n-1}< (n^{2}+\frac{n+1}{2})^{2}
Where have the expressions for the boundaries come from? Are they related to n=3 in some way? Because if not, one could also say that this holds for n > 2 or n > 5 etc.
Basically, whats the relationship between the boundaries used and n > 3?
Aquafina,
Take fardeen's 'n' even case, expand the equation $\displaystyle (n^{2}+\frac{n}{2})^{2}- \frac{n^{5}-1}{n-1}$ and tell me what you get(is it positive or negative?). Then do the same thing with $\displaystyle (n^{2}+\frac{n}{2}+1)^{2} - \frac{n^{5}-1}{n-1} $.
Can you try it? Then tell us if you are stuck somewhere...
I get $\displaystyle -(\frac34 n^2 + n + 1)$
I asked you to compute the difference:and the second part to
n^4 + n^3 + 9/4 n^2 + n + 1
$\displaystyle n^4 + n^3 + 9/4 n^2 + n + 1 - (n^4+n^3+n^2+n+1) = \frac{5}{4}n^2$
Next step:
Prove that $\displaystyle -(\frac34 n^2 + n + 1) < 0$ and $\displaystyle \frac{5}{4}n^2 > 0$
$\displaystyle \frac{5}{4}n^2 > 0$ is obvious
$\displaystyle -(\frac34 n^2 + n + 1) < 0$ is kinda obvious too..
So now try the odd case
Probably 9 was a typo, good work:
I get $\displaystyle (n^{2}+\frac{n-1}{2})^{2} - \frac{n^{5}-1}{n-1} = -\left(\frac74 n^2 + \frac32 n + \frac34\right) < 0$
Second one is right: $\displaystyle (n^{2}+\frac{n+1}{2})^{2}- \frac{n^{5}-1}{n-1} = \frac{n^2 - 2n - 3}{4}$. So when do you think $\displaystyle n^2 - 2n - 3 > 0 $ holds?
ah that's really good! (yes 9 was a typo sorry)
how come only the upper boundary of n being odd is related to n > 3? i.e. with n^2 - 2n - 3 > 0
Also, how would you come up with these equations from scratch!? I get how I worked back from fardeen's work to get the case...