Find sum of all natural numbers $n $ such that $(n^2+n+1)^2$ divides $1+n+n^2+...+n^{2195}$.

Question

Update

Find sum of all $n \in \mathbb{N}$ such that $(n^2+n+1)^2$ divides $n^{2195}+n^{2194}+...+n^2+n+1$.

I have no idea. Can anyone help? At least with a hint. Thanks for the help in advance.

Original Question (Before Correction)

Find sum of all natural numbers $n $ such that $n^2+n+1$ divides $1+n+n^2+...+n^{2195}$.

for $n=1$, trivial. then, how to move to the next step for $n≥2$ ? — lap lapan, Apr 12 '20 at 10:58
The question is contradictory. It says "find all natural numbers $n$" and then "find the sum of all $n$". — Dietrich Burde, Apr 12 '20 at 10:59
Have you tried it for smaller values of $2195$? Oh, and \cdots gives you a nicer ellipsis. — Angina Seng, Apr 12 '20 at 11:00
Might be easier to think about $n^3-1$ dividing $n^{2196}-1$. — Gerry Myerson, Apr 12 '20 at 11:01
@ Dietrich Burde Isn't it possible to factor out $n^2+n+1$ from the large polynomial? — Nεo Pλατo, Apr 12 '20 at 11:03
@Plato: That's just what I was thinking. So the condition is true for all $n\in \Bbb N$. Arnold, are you sure you have it right? — TonyK, Apr 12 '20 at 11:06
@arnold If you change the question after getting answers, you should address it. It is not appropriate to change your question, rendering previous answers irrelevant. In fact, it is a better idea to create a new question, rather than changing the question. And you should also notify the authors of old answers of the changes. — Batominovski, Apr 12 '20 at 15:37
Can't answer right now, but the solutions are 1 and 13 (and 0 if you consider it a natural number, but that doesn't change the sum). — aschepler, Apr 12 '20 at 19:55

score 1 · Answer 1 · answered Apr 12 '20 at 11:17

1

Here's hint: this sum is equal to $$\frac{n^{2196}-1}{n-1}=\frac{\bigl(n^3\bigr)^{732}-1}{n-1}=\dotsm$$

answered Apr 12 '20 at 11:17

Bernard

175,478

Nεo Pλατo · Answer 2 · 2020-04-12T19:40:38.227

1

As Gerry Myerson has stated in the comments, it just might be helpful. The underlying fact is that it's still divisible nevertheless. Watch this.

Say $n^3-1=k$. Then we want to evaluate $(n^{2196}-1) \bmod k$

$n^3=k+1$

$\begin{align} (n^3)^{732}-1 &= (k+1)^{732}-1 \\ &\equiv 1-1 \pmod{k} \\ &\equiv 0 \pmod{k} \end{align}$

So it is divisible for all $n \in \Bbb N$, which by the factoring idea I presented in the comments still holds.

Edit: Let's try out the case I've been presented with: the big polynomial versus $(n^2+n+1)^2$

So the polynomial can be factored into:

$(n^2+n+1)\color{red}{(1+n^3+n^6+n^9+\ldots+n^{2193})}$

So the question now really is how the red factor can be divided by $n^2+n+1$. It's value is:

$\begin{align}S&=1+\color{blue}{n^3+n^6+n^9 \ldots n^{2193}} \\ n^3 S&=\color{blue}{n^3+n^6 \ldots n^{2193}} + n^{2196} \\ n^3 S - S&= n^{2196}-1 \\ S&=\dfrac{n^{2196}-1}{n^3-1} \end{align}$

So what we're checking for is:

$\dfrac{\dfrac{n^{2196}-1}{n^3-1}}{\dfrac{n^3-1}{n-1}}=\dfrac{(n-1)(n^{2196}-1)}{(n^3-1)^2}$

And since:

$n^{2196}-1=(n^3-1)\color{red}{(n^{2193}+n^{2190} \ldots 1)}$

So what we're looking at now is the red sum versus $n^3-1$. Doing the same thing as before with $n^3=k+1$:

$=(n^3)^{731}+(n^3)^{730}+(n^3)^{729} \ldots n^3+1$

$=(k+1)^{731}+(k+1)^{730} \ldots (k+1)+1$

$\equiv 732 \pmod k$

So $(n^3-1)$ must divide $732$ meaning that one of it's factors should be one less than a perfect cube.

Factors of $732$ plus one are:

$\{2,3,5,7,12,62,123,184,245,367,733\}$

None of them are perfect cubes so you can conclude the only solutions are the trivial ones which are $0$ and $1$.

edited Apr 12 '20 at 19:40

answered Apr 12 '20 at 11:23

Nεo Pλατo

1,705

what about $(n^2+n+1)^2$ divides $n^{2195}+...+n+1$ ? – lap lapan Apr 12 '20 at 11:29
if $n^2+n+1$ divides $n^{2195}+...+n+1$, then $(n^2+n+1)^2$ so? – lap lapan Apr 12 '20 at 11:31
I'll get back to you when I figure it out. – Nεo Pλατo Apr 12 '20 at 12:07
How's my answer now? – Nεo Pλατo Apr 12 '20 at 14:14
well, i got the sense now. great! – lap lapan Apr 12 '20 at 14:55
Sure thing. But don't take this answer as a standard because I feel like there's a quicker way to do this. – Nεo Pλατo Apr 12 '20 at 16:05
What happened to the $(n-1)$ in the quotient after "what we're checking for is"? Near the end, you seem to mean "$(n^3-1)$ must divide 732" instead of "732 must divide $(n^3-1)$". – aschepler Apr 12 '20 at 18:46
@aschepler The $n-1$ is a numerator so it doesn't worry us much. As for the divides thing, I always confuse the words. I'll change it right now – Nεo Pλατo Apr 12 '20 at 19:43
So with $a=(n^3-1)^2$, $b=n-1$, $c=n^{2196}-1$ you're saying "if $a$ divides $bc$, then $a$ divides $c$". – aschepler Apr 12 '20 at 20:00
More accurate that if $c$ divides $b$ then it also divides $ab$. That's why I didn't consider $a$ – Nεo Pλατo Apr 12 '20 at 20:09
Sure, so if you found $n$ such that $a$ divides $c$, then you also know $a$ divides $bc$ and you've found some solutions. But then it seems you're concluding since $a$ does not divide $c$ for $n>1$, then $a$ does not divide $bc$. – aschepler Apr 12 '20 at 22:05
Is that wrong? I'm only a beginner in such things so I'd appreciate advice. Nevertheless, I ran this on Wolfram Alpha and this time I'm safe – Nεo Pλατo Apr 12 '20 at 22:22
Speaking of Wolfram Alpha, it evaluates $((13^{2196} - 1)/12) \bmod 183$ as $0$. So $n=13$ is also a solution, which means your logic is flawed. – TonyK Apr 14 '20 at 12:00
Wait a minute, that number $183$ was in my answer. Let me redo – Nεo Pλατo Apr 14 '20 at 12:01
Is the number divisible by $183^2$ because that was the question asked :) – Nεo Pλατo Apr 14 '20 at 12:05
@Plato: Yes it is. Sorry, my mistake. – TonyK Apr 16 '20 at 09:10
@TonyK Either way I'm still half right. Thanks for the help – Nεo Pλατo Apr 16 '20 at 09:13

Empy2 · Accepted Answer · 2020-04-12T15:03:23.170

1

$$n^{2193}+n^{2190}+\cdots+1 = n^{2193}-1+n^{2190}-1+\cdots+1-1+732$$ The left-hand side is the result of dividing by the first factor of $n^2+n+1$. The $n^{3k}-1$ are each multiples of the second factor of $n^2+n+1$, leaving $732$ alone.
$n^2+n+1$ is odd, and odd factors of 732 are $1,3,61,183$.
$$1=0^2+0+1\\3=1^2+1+1\\183=13^2+13+1$$

edited Apr 12 '20 at 15:03

answered Apr 12 '20 at 12:27

Empy2

50,853

mmm.. sorry? i can't understand it. can you explain? – lap lapan Apr 12 '20 at 12:36
Although I understand what this equality means, it might be better if you could elaborate a bit. From review – Ѕᴀᴀᴅ Apr 12 '20 at 13:09
what if $(n^2+n+1)^2$ ? – lap lapan Apr 12 '20 at 22:52

Find sum of all natural numbers $n $ such that $(n^2+n+1)^2$ divides $1+n+n^2+...+n^{2195}$.

3 Answers3