1. ## Proof by Induction?

moved to a new thread for general case

2. ## Difficult Problem

I have not had luck with induction here. However, here are two very unusual observations I have made:

$\lim_{n \rightarrow \infty} frac( \sqrt{\frac{n^5-1}{n-1}}) = \frac{3}{8}$ for n even and $\frac{7}{8}$ for n odd. In other words, the decimal expansion seems to be tending towards an alternating pattern between .375 and .875. To be a perfect square, this decimal would of course need to be exactly zero, so proving this observation would imply your hypothesis.

Since it does alternate, you may try using induction where $n \rightarrow n+2$

And don't forget the identity $\frac{n^5-1}{n-1}=n^4+n^3+n^2+n+1$

3. Hi, might seem like a stupid q but go should I prove that using induction?

So far as I know, induction is not effective for proving a limit exists. Here is what I have gotten so far:

Let $f(n)=n^4+n^3+n^2+n+1$

As n gets large, $\sqrt{f(n+2)}-\sqrt{f(n)} \approx 4n+5$

More rigorously, $\lim_{n \rightarrow \infty} \frac{\sqrt{f(n+2)}-\sqrt{f(n)}}{4n+5}=1$

The actual value $4n+5$ is not so important as the fact that it is an integer. What this essentially means is that if $f(n)$ is not a perfect square, then $f(n+2)$ won't be either, since both of their square roots will have increasingly identical decimal expansions $\approx .375$ or $.875$.

You can actually graph $\sqrt{f(n+2)}-\sqrt{f(n)}$ and $4n+5$ and see how astonishingly closely they overlap. I am currently looking around for a method of taking the square root of a polynomial. I'm pretty sure it exists, basically given $f(x)=Ax^4+Bx^3+Cx^2+Dx+E$ , find coefficients $a_n$ that satisfy $f(x)=(a_0x^2+a_1x+a_2+a_3x^{-1}+a_4x^{-2}+...)^2$ , effectively finding the square root of a polynomial.

I believe that rewriting $\sqrt{f(n+2)}-\sqrt{f(n)}$ as series, a bunch of stuff will cancel, leaving only $4n+5$ and some extra term that goes to zero fairly quickly.

I have the sneaking suspicion that I am using a cannon to kill a mosquito, and a super member will come along with a painfully obvious elementary proof.

5. i'll try that method thanks

any 'super member' u have in mind that i could PM to ask for any existing simpler method?

6. Here are some guys that have put me to shame in other posts: chisigma, Aryth, PaulRS, fardeen_gen, CaptainBlack, Gamma, Soroban. Not sure what their particular areas of interest are, though.

7. Thanks, I PM-ed them. Hopefully one of them has an idea

8. Originally Posted by Aquafina
Hi, I need to show that for n > 3:

(n^5 - 1)/(n-1) ≠ k^2 (does not equal)

for any integer 'k'

i.e. it is not a perfect square
For $n > 3$

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,
$(n^{2}+\frac{n-1}{2})^{2}<\frac{n^{5}-1}{n-1}< (n^{2}+\frac{n+1}{2})^{2}$

If the above statements are true,
the value of $\frac{n^{5}-1}{n-1}$ is between two consecutive squares no matter $n$ is even or odd $\implies \frac{n^{5}-1}{n-1}\neq k^{2}$, for any integer $k$.

For verification,

$n = 4\ \Rightarrow\ (18)^2 < 341 < (19)^2$.
$n = 5\ \Rightarrow\ (27)^2 < 781 < (28)^2$

I think we are done then!

9. thanks

i had a question, how does this prove it for n>3

because when n=3, k^2 = 121 which is a perfect square

so how come only when n>3, the proof you showed should work?

10. Originally Posted by Aquafina
thanks

i had a question, how does this prove it for n<=3

because when n=3, k^2 = 121 which is a perfect square

so how come only when n>3, the proof you showed should work?
Originally Posted by Aquafina
Hi, I need to show that for n > 3...

The statement is only valid for $n > 3$ and not $n\leq 3$. The question mentions it too

11. yeah but the original question was to find any values of n for which

n^4+n^3+n^2+n+1 is a perfect square

i found out that it works for n=3, and guessed that there would be no other value, using excel for very large value of n!

so i decided to prove

(n^5 - 1)/(n-1) ≠ k^2 for n>3

12. Originally Posted by Aquafina
so i decided to prove

(n^5 - 1)/(n-1) ≠ k^2 for n>3
So, what you decided to prove has been proved, right?

13. haha yes, but can the argument you gave be used towards the original question?

14. The original question, as far as I am concerned is:
Hi, I need to show that for n > 3:

(n^5 - 1)/(n-1) ≠ k^2 (does not equal)

for any integer 'k'

i.e. it is not a perfect square

And of course my argument only holds for the proof you wanted.(make that "initially" wanted )

15. lol ok, will this proof imply that n=3 is the only solution for n^4+n^3+n^2+n+1 being a perfect square?

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