1. Originally Posted by Aquafina
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?

Yes!

2. ok thanks

now do you have any idea how i can show that

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

for ONLY n=3

where k is an integer

3. Originally Posted by Aquafina
ok thanks

now do you have any idea how i can show that

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

for ONLY n=3

where k is an integer
?

4. sorry, does equal k^2

5. The only way I see, is to prove that $\frac{n^5 - 1}{n - 1}\neq k^2$ for $n\neq 3$. We did the $n > 3$ part. I think you can make a separate thread for $n < 2$.

6. Originally Posted by fardeen_gen
The only way I see, is to prove that $\frac{n^5 - 1}{n - 1}\neq k^2$ for $n\neq 3$. We did the $n > 3$ part. I think you can make a separate thread for $n < 2$.
I thought that n is a natural number... Because then by verifying for n = 1,2 you can see the expression's not a square. However n > 3 case, we already know the result. Thus n=3 is the only number that makes the expression a square...

7. 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?

8. Originally Posted by Aquafina
Basically, whats the relationship between the boundaries used and n > 3?
Did you expand fardeen's ingenious inequalities and see why they are true?

9. sorry ill just paste the expression again in picture format

CASE I: is even,
<font size="3">

i wasnt able to see the relationship sorry lol..

i'm not an IIT student, i'm guessing u guys are

could you tell me how they help?

10. Originally Posted by Aquafina
sorry ill just paste the expression again in picture format

CASE I: is even,
<font size="3">

i wasnt able to see the relationship sorry lol..

i'm not an IIT student, i'm guessing u guys are

could you tell me how they help?
Aquafina,

Take fardeen's 'n' even case, expand the equation $(n^{2}+\frac{n}{2})^{2}- \frac{n^{5}-1}{n-1}$ and tell me what you get(is it positive or negative?). Then do the same thing with $(n^{2}+\frac{n}{2}+1)^{2} - \frac{n^{5}-1}{n-1}$.

Can you try it? Then tell us if you are stuck somewhere...

11. ok i changed (n^5 - 1)/(n-1) to n^4 + n^3 + n^2 + n + 1

then for the first part:

-3/4 n^2 + n + 1

and the second part to

n^4 + n^3 + 9/4 n^2 + n + 1

12. Originally Posted by Aquafina
ok i changed (n^5 - 1)/(n-1) to n^4 + n^3 + n^2 + n + 1

then for the first part:

-3/4 n^2 + n + 1
I get $-(\frac34 n^2 + n + 1)$

and the second part to

n^4 + n^3 + 9/4 n^2 + n + 1
I asked you to compute the difference:
$n^4 + n^3 + 9/4 n^2 + n + 1 - (n^4+n^3+n^2+n+1) = \frac{5}{4}n^2$

Next step:
Prove that $-(\frac34 n^2 + n + 1) < 0$ and $\frac{5}{4}n^2 > 0$

$\frac{5}{4}n^2 > 0$ is obvious

$-(\frac34 n^2 + n + 1) < 0$ is kinda obvious too..

So now try the odd case

13. okay..

- (7/9 n^2 + 3/2 n + 3/4)

and

1/4 n^2 - 1/2 n - 3/4

14. Originally Posted by Aquafina
okay..

- (7/9 n^2 + 3/2 n + 3/4)

and

1/4 n^2 - 1/2 n - 3/4
Probably 9 was a typo, good work:

I get $(n^{2}+\frac{n-1}{2})^{2} - \frac{n^{5}-1}{n-1} = -\left(\frac74 n^2 + \frac32 n + \frac34\right) < 0$

Second one is right: $(n^{2}+\frac{n+1}{2})^{2}- \frac{n^{5}-1}{n-1} = \frac{n^2 - 2n - 3}{4}$. So when do you think $n^2 - 2n - 3 > 0$ holds?

15. 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...

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