Find all natural numbers $n$, such that polynomial $n^7+n^6+n^5+1$ would have exactly 3 divisors.
$begingroup$
Find all natural numbers $n$, such that polynomial $$n^7+n^6+n^5+1$$ would have exactly 3 divisors.
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors. Also, n must be an even number, otherwise this polynomial would be divisible by 4 (if it's an odd number, then it would be possible and would fit conditions only and only if n=1). My guts are telling me that somehow I have to prove that it's impossible with even numbers too, though I have no idea how.
number-theory elementary-number-theory polynomials
edited Dec 2 '18 at 19:32
greedoid
38.7k114797
asked Dec 2 '18 at 18:36
Severus156Severus156
1637
$endgroup$
add a comment |
$begingroup$
Find all natural numbers $n$, such that polynomial $$n^7+n^6+n^5+1$$ would have exactly 3 divisors.
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors. Also, n must be an even number, otherwise this polynomial would be divisible by 4 (if it's an odd number, then it would be possible and would fit conditions only and only if n=1). My guts are telling me that somehow I have to prove that it's impossible with even numbers too, though I have no idea how.
number-theory elementary-number-theory polynomials
edited Dec 2 '18 at 19:32
greedoid
38.7k114797
asked Dec 2 '18 at 18:36
Severus156Severus156
1637
$endgroup$
add a comment |
$begingroup$
Find all natural numbers $n$, such that polynomial $$n^7+n^6+n^5+1$$ would have exactly 3 divisors.
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors. Also, n must be an even number, otherwise this polynomial would be divisible by 4 (if it's an odd number, then it would be possible and would fit conditions only and only if n=1). My guts are telling me that somehow I have to prove that it's impossible with even numbers too, though I have no idea how.
number-theory elementary-number-theory polynomials
edited Dec 2 '18 at 19:32
greedoid
38.7k114797
asked Dec 2 '18 at 18:36
Severus156Severus156
1637
$endgroup$
Find all natural numbers $n$, such that polynomial $$n^7+n^6+n^5+1$$ would have exactly 3 divisors.
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors. Also, n must be an even number, otherwise this polynomial would be divisible by 4 (if it's an odd number, then it would be possible and would fit conditions only and only if n=1). My guts are telling me that somehow I have to prove that it's impossible with even numbers too, though I have no idea how.
number-theory elementary-number-theory polynomials
number-theory elementary-number-theory polynomials
edited Dec 2 '18 at 19:32
greedoid
38.7k114797
asked Dec 2 '18 at 18:36
Severus156Severus156
1637
edited Dec 2 '18 at 19:32
greedoid
38.7k114797
asked Dec 2 '18 at 18:36
Severus156Severus156
1637
edited Dec 2 '18 at 19:32
greedoid
38.7k114797
edited Dec 2 '18 at 19:32
greedoid
38.7k114797
edited Dec 2 '18 at 19:32
greedoid
38.7k114797
38.7k114797
asked Dec 2 '18 at 18:36
Severus156Severus156
1637
asked Dec 2 '18 at 18:36
Severus156Severus156
1637
asked Dec 2 '18 at 18:36
Severus156Severus156
1637
1637
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can factorize this polynomial like this: $$(n+1)(n^2+1)(n^4-n+1)$$ so this one has almost always at least 3 divisors.
Actualy, for $n>1$ we have $n+1>2$, $n^2+1>2$ and $n^4-n+1>2$ so it could be only $n=1$ which works.
How I got this this factorization:
begin{eqnarray} n^7+n^6+n^5+1 &=& n^5(n^2+1) + (n^2+1)(n^4-n^2+1)\
&=& (n^2+1)(n^5+n^4-n^2+1)\
&=& (n^2+1)Big(n^4(n+1)-(n-1)(n+1)Big)\
&=&(n+1)(n^2+1)(n^4-n+1)
end{eqnarray}
answered Dec 2 '18 at 18:41
greedoidgreedoid
38.7k114797
$endgroup$
$begingroup$
Seems approvable. And also I'm quite interested - how do you create such hard factorizations for such untraditional polynomials? Is there any secret or are you just using some programs/apps for this?
$endgroup$
– Severus156
Dec 2 '18 at 18:49
1
$begingroup$
@Severus156 You sometimes see a root easily; like in this case: -1 is an easy to find root. So you can divide the polynomial by $n+1$. Then repeat the process.
$endgroup$
– Guus Palmer
Dec 2 '18 at 18:51
1
$begingroup$
But further roots are not as beautiful as before, $n^2+1$ root is $i$, so basically you just guess every single integer close to $0$ and hope that you will find those roots, is that so @GuusPalmer?
$endgroup$
– Severus156
Dec 2 '18 at 18:54
1
$begingroup$
@Severus156 That's what I would do most of the time. But you should also be aware that Number Theory is full of tricks; i.e. there's not always a general recipe for everything, just look at Mordell equations: they use a large variety of methods to solve these. To not wander away from the subject too much, notice that the polynomial has an even degree and two unevens in it. That sometimes indicates a $pm1$ or $pm i$...
$endgroup$
– Guus Palmer
Dec 2 '18 at 19:00
$begingroup$
"Actualy, for n>1 we have n+1>2, n2+1>2 and n4−n+1>2 so it could be only n=1 which works" Why is that relevant? We will have exactly 3 factors if all $n+1$ and $n^2 + 1$ and $n^4 - n + 1$ are prime.. Why can't that happen if $n > 1$?
$endgroup$
– fleablood
Dec 2 '18 at 20:26
|
show 3 more comments
$begingroup$
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors.
Or fewer.
To rephrase and elaborate on greedoid's answer in a way that doesn't involve complete factoring:
So if if the polynomial factor as $r(n)q(n)$ we must have either $r(n) = 1; q(n) = p^2$ for some $n$ and prime $p$ or $r(n) =q(n)=p$ for some $n$ and prime $p$.
Note that if $n =-1$ we get the polynomial is $0$ so $n+1$ factors.
And $frac {n^7 + n^6 + n^5 + 1}{n+1} = n^6 + n^4 -n^3 +n^2 -n + 1$.
If $n+1 = 1$ then we get $n = 0$ and .... the other is $1$ and that won't do.
It's pretty clear if $n > 1$ then $n^6 + ... etc > 1$ so $(n+1) = p^2$ and $n^6 + ... etc = 1$ is out of the question.
That just leaves $n + 1 = n^6 + .... = p$ as an option. And for that to work we must have $n = 1$. And in that case it does work out that $n+1 = n^6 + ... etc = 2$.
answered Dec 2 '18 at 21:23
fleabloodfleablood
68.7k22685
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can factorize this polynomial like this: $$(n+1)(n^2+1)(n^4-n+1)$$ so this one has almost always at least 3 divisors.
Actualy, for $n>1$ we have $n+1>2$, $n^2+1>2$ and $n^4-n+1>2$ so it could be only $n=1$ which works.
How I got this this factorization:
begin{eqnarray} n^7+n^6+n^5+1 &=& n^5(n^2+1) + (n^2+1)(n^4-n^2+1)\
&=& (n^2+1)(n^5+n^4-n^2+1)\
&=& (n^2+1)Big(n^4(n+1)-(n-1)(n+1)Big)\
&=&(n+1)(n^2+1)(n^4-n+1)
end{eqnarray}
answered Dec 2 '18 at 18:41
greedoidgreedoid
38.7k114797
$endgroup$
$begingroup$
Seems approvable. And also I'm quite interested - how do you create such hard factorizations for such untraditional polynomials? Is there any secret or are you just using some programs/apps for this?
$endgroup$
– Severus156
Dec 2 '18 at 18:49
1
$begingroup$
@Severus156 You sometimes see a root easily; like in this case: -1 is an easy to find root. So you can divide the polynomial by $n+1$. Then repeat the process.
$endgroup$
– Guus Palmer
Dec 2 '18 at 18:51
1
$begingroup$
But further roots are not as beautiful as before, $n^2+1$ root is $i$, so basically you just guess every single integer close to $0$ and hope that you will find those roots, is that so @GuusPalmer?
$endgroup$
– Severus156
Dec 2 '18 at 18:54
1
$begingroup$
@Severus156 That's what I would do most of the time. But you should also be aware that Number Theory is full of tricks; i.e. there's not always a general recipe for everything, just look at Mordell equations: they use a large variety of methods to solve these. To not wander away from the subject too much, notice that the polynomial has an even degree and two unevens in it. That sometimes indicates a $pm1$ or $pm i$...
$endgroup$
– Guus Palmer
Dec 2 '18 at 19:00
$begingroup$
"Actualy, for n>1 we have n+1>2, n2+1>2 and n4−n+1>2 so it could be only n=1 which works" Why is that relevant? We will have exactly 3 factors if all $n+1$ and $n^2 + 1$ and $n^4 - n + 1$ are prime.. Why can't that happen if $n > 1$?
$endgroup$
– fleablood
Dec 2 '18 at 20:26
|
show 3 more comments
$begingroup$
You can factorize this polynomial like this: $$(n+1)(n^2+1)(n^4-n+1)$$ so this one has almost always at least 3 divisors.
Actualy, for $n>1$ we have $n+1>2$, $n^2+1>2$ and $n^4-n+1>2$ so it could be only $n=1$ which works.
How I got this this factorization:
begin{eqnarray} n^7+n^6+n^5+1 &=& n^5(n^2+1) + (n^2+1)(n^4-n^2+1)\
&=& (n^2+1)(n^5+n^4-n^2+1)\
&=& (n^2+1)Big(n^4(n+1)-(n-1)(n+1)Big)\
&=&(n+1)(n^2+1)(n^4-n+1)
end{eqnarray}
answered Dec 2 '18 at 18:41
greedoidgreedoid
38.7k114797
$endgroup$
$begingroup$
Seems approvable. And also I'm quite interested - how do you create such hard factorizations for such untraditional polynomials? Is there any secret or are you just using some programs/apps for this?
$endgroup$
– Severus156
Dec 2 '18 at 18:49
1
$begingroup$
@Severus156 You sometimes see a root easily; like in this case: -1 is an easy to find root. So you can divide the polynomial by $n+1$. Then repeat the process.
$endgroup$
– Guus Palmer
Dec 2 '18 at 18:51
1
$begingroup$
But further roots are not as beautiful as before, $n^2+1$ root is $i$, so basically you just guess every single integer close to $0$ and hope that you will find those roots, is that so @GuusPalmer?
$endgroup$
– Severus156
Dec 2 '18 at 18:54
1
$begingroup$
@Severus156 That's what I would do most of the time. But you should also be aware that Number Theory is full of tricks; i.e. there's not always a general recipe for everything, just look at Mordell equations: they use a large variety of methods to solve these. To not wander away from the subject too much, notice that the polynomial has an even degree and two unevens in it. That sometimes indicates a $pm1$ or $pm i$...
$endgroup$
– Guus Palmer
Dec 2 '18 at 19:00
$begingroup$
"Actualy, for n>1 we have n+1>2, n2+1>2 and n4−n+1>2 so it could be only n=1 which works" Why is that relevant? We will have exactly 3 factors if all $n+1$ and $n^2 + 1$ and $n^4 - n + 1$ are prime.. Why can't that happen if $n > 1$?
$endgroup$
– fleablood
Dec 2 '18 at 20:26
|
show 3 more comments
$begingroup$
You can factorize this polynomial like this: $$(n+1)(n^2+1)(n^4-n+1)$$ so this one has almost always at least 3 divisors.
Actualy, for $n>1$ we have $n+1>2$, $n^2+1>2$ and $n^4-n+1>2$ so it could be only $n=1$ which works.
How I got this this factorization:
begin{eqnarray} n^7+n^6+n^5+1 &=& n^5(n^2+1) + (n^2+1)(n^4-n^2+1)\
&=& (n^2+1)(n^5+n^4-n^2+1)\
&=& (n^2+1)Big(n^4(n+1)-(n-1)(n+1)Big)\
&=&(n+1)(n^2+1)(n^4-n+1)
end{eqnarray}
answered Dec 2 '18 at 18:41
greedoidgreedoid
38.7k114797
$endgroup$
You can factorize this polynomial like this: $$(n+1)(n^2+1)(n^4-n+1)$$ so this one has almost always at least 3 divisors.
Actualy, for $n>1$ we have $n+1>2$, $n^2+1>2$ and $n^4-n+1>2$ so it could be only $n=1$ which works.
How I got this this factorization:
begin{eqnarray} n^7+n^6+n^5+1 &=& n^5(n^2+1) + (n^2+1)(n^4-n^2+1)\
&=& (n^2+1)(n^5+n^4-n^2+1)\
&=& (n^2+1)Big(n^4(n+1)-(n-1)(n+1)Big)\
&=&(n+1)(n^2+1)(n^4-n+1)
end{eqnarray}
answered Dec 2 '18 at 18:41
greedoidgreedoid
38.7k114797
edited Dec 2 '18 at 18:52
answered Dec 2 '18 at 18:41
greedoidgreedoid
38.7k114797
answered Dec 2 '18 at 18:41
greedoidgreedoid
38.7k114797
answered Dec 2 '18 at 18:41
greedoidgreedoid
38.7k114797
38.7k114797
$begingroup$
Seems approvable. And also I'm quite interested - how do you create such hard factorizations for such untraditional polynomials? Is there any secret or are you just using some programs/apps for this?
$endgroup$
– Severus156
Dec 2 '18 at 18:49
1
$begingroup$
@Severus156 You sometimes see a root easily; like in this case: -1 is an easy to find root. So you can divide the polynomial by $n+1$. Then repeat the process.
$endgroup$
– Guus Palmer
Dec 2 '18 at 18:51
1
$begingroup$
But further roots are not as beautiful as before, $n^2+1$ root is $i$, so basically you just guess every single integer close to $0$ and hope that you will find those roots, is that so @GuusPalmer?
$endgroup$
– Severus156
Dec 2 '18 at 18:54
1
$begingroup$
@Severus156 That's what I would do most of the time. But you should also be aware that Number Theory is full of tricks; i.e. there's not always a general recipe for everything, just look at Mordell equations: they use a large variety of methods to solve these. To not wander away from the subject too much, notice that the polynomial has an even degree and two unevens in it. That sometimes indicates a $pm1$ or $pm i$...
$endgroup$
– Guus Palmer
Dec 2 '18 at 19:00
$begingroup$
"Actualy, for n>1 we have n+1>2, n2+1>2 and n4−n+1>2 so it could be only n=1 which works" Why is that relevant? We will have exactly 3 factors if all $n+1$ and $n^2 + 1$ and $n^4 - n + 1$ are prime.. Why can't that happen if $n > 1$?
$endgroup$
– fleablood
Dec 2 '18 at 20:26
|
show 3 more comments
$begingroup$
Seems approvable. And also I'm quite interested - how do you create such hard factorizations for such untraditional polynomials? Is there any secret or are you just using some programs/apps for this?
$endgroup$
– Severus156
Dec 2 '18 at 18:49
1
$begingroup$
@Severus156 You sometimes see a root easily; like in this case: -1 is an easy to find root. So you can divide the polynomial by $n+1$. Then repeat the process.
$endgroup$
– Guus Palmer
Dec 2 '18 at 18:51
1
$begingroup$
But further roots are not as beautiful as before, $n^2+1$ root is $i$, so basically you just guess every single integer close to $0$ and hope that you will find those roots, is that so @GuusPalmer?
$endgroup$
– Severus156
Dec 2 '18 at 18:54
1
$begingroup$
@Severus156 That's what I would do most of the time. But you should also be aware that Number Theory is full of tricks; i.e. there's not always a general recipe for everything, just look at Mordell equations: they use a large variety of methods to solve these. To not wander away from the subject too much, notice that the polynomial has an even degree and two unevens in it. That sometimes indicates a $pm1$ or $pm i$...
$endgroup$
– Guus Palmer
Dec 2 '18 at 19:00
$begingroup$
"Actualy, for n>1 we have n+1>2, n2+1>2 and n4−n+1>2 so it could be only n=1 which works" Why is that relevant? We will have exactly 3 factors if all $n+1$ and $n^2 + 1$ and $n^4 - n + 1$ are prime.. Why can't that happen if $n > 1$?
$endgroup$
– fleablood
Dec 2 '18 at 20:26
$begingroup$
Seems approvable. And also I'm quite interested - how do you create such hard factorizations for such untraditional polynomials? Is there any secret or are you just using some programs/apps for this?
$endgroup$
– Severus156
Dec 2 '18 at 18:49
$begingroup$
Seems approvable. And also I'm quite interested - how do you create such hard factorizations for such untraditional polynomials? Is there any secret or are you just using some programs/apps for this?
$endgroup$
– Severus156
Dec 2 '18 at 18:49
1
1
$begingroup$
@Severus156 You sometimes see a root easily; like in this case: -1 is an easy to find root. So you can divide the polynomial by $n+1$. Then repeat the process.
$endgroup$
– Guus Palmer
Dec 2 '18 at 18:51
$begingroup$
@Severus156 You sometimes see a root easily; like in this case: -1 is an easy to find root. So you can divide the polynomial by $n+1$. Then repeat the process.
$endgroup$
– Guus Palmer
Dec 2 '18 at 18:51
1
1
$begingroup$
But further roots are not as beautiful as before, $n^2+1$ root is $i$, so basically you just guess every single integer close to $0$ and hope that you will find those roots, is that so @GuusPalmer?
$endgroup$
– Severus156
Dec 2 '18 at 18:54
$begingroup$
But further roots are not as beautiful as before, $n^2+1$ root is $i$, so basically you just guess every single integer close to $0$ and hope that you will find those roots, is that so @GuusPalmer?
$endgroup$
– Severus156
Dec 2 '18 at 18:54
1
1
$begingroup$
@Severus156 That's what I would do most of the time. But you should also be aware that Number Theory is full of tricks; i.e. there's not always a general recipe for everything, just look at Mordell equations: they use a large variety of methods to solve these. To not wander away from the subject too much, notice that the polynomial has an even degree and two unevens in it. That sometimes indicates a $pm1$ or $pm i$...
$endgroup$
– Guus Palmer
Dec 2 '18 at 19:00
$begingroup$
@Severus156 That's what I would do most of the time. But you should also be aware that Number Theory is full of tricks; i.e. there's not always a general recipe for everything, just look at Mordell equations: they use a large variety of methods to solve these. To not wander away from the subject too much, notice that the polynomial has an even degree and two unevens in it. That sometimes indicates a $pm1$ or $pm i$...
$endgroup$
– Guus Palmer
Dec 2 '18 at 19:00
$begingroup$
"Actualy, for n>1 we have n+1>2, n2+1>2 and n4−n+1>2 so it could be only n=1 which works" Why is that relevant? We will have exactly 3 factors if all $n+1$ and $n^2 + 1$ and $n^4 - n + 1$ are prime.. Why can't that happen if $n > 1$?
$endgroup$
– fleablood
Dec 2 '18 at 20:26
$begingroup$
"Actualy, for n>1 we have n+1>2, n2+1>2 and n4−n+1>2 so it could be only n=1 which works" Why is that relevant? We will have exactly 3 factors if all $n+1$ and $n^2 + 1$ and $n^4 - n + 1$ are prime.. Why can't that happen if $n > 1$?
$endgroup$
– fleablood
Dec 2 '18 at 20:26
|
show 3 more comments
$begingroup$
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors.
Or fewer.
To rephrase and elaborate on greedoid's answer in a way that doesn't involve complete factoring:
So if if the polynomial factor as $r(n)q(n)$ we must have either $r(n) = 1; q(n) = p^2$ for some $n$ and prime $p$ or $r(n) =q(n)=p$ for some $n$ and prime $p$.
Note that if $n =-1$ we get the polynomial is $0$ so $n+1$ factors.
And $frac {n^7 + n^6 + n^5 + 1}{n+1} = n^6 + n^4 -n^3 +n^2 -n + 1$.
If $n+1 = 1$ then we get $n = 0$ and .... the other is $1$ and that won't do.
It's pretty clear if $n > 1$ then $n^6 + ... etc > 1$ so $(n+1) = p^2$ and $n^6 + ... etc = 1$ is out of the question.
That just leaves $n + 1 = n^6 + .... = p$ as an option. And for that to work we must have $n = 1$. And in that case it does work out that $n+1 = n^6 + ... etc = 2$.
answered Dec 2 '18 at 21:23
fleabloodfleablood
68.7k22685
$endgroup$
add a comment |
$begingroup$
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors.
Or fewer.
To rephrase and elaborate on greedoid's answer in a way that doesn't involve complete factoring:
So if if the polynomial factor as $r(n)q(n)$ we must have either $r(n) = 1; q(n) = p^2$ for some $n$ and prime $p$ or $r(n) =q(n)=p$ for some $n$ and prime $p$.
Note that if $n =-1$ we get the polynomial is $0$ so $n+1$ factors.
And $frac {n^7 + n^6 + n^5 + 1}{n+1} = n^6 + n^4 -n^3 +n^2 -n + 1$.
If $n+1 = 1$ then we get $n = 0$ and .... the other is $1$ and that won't do.
It's pretty clear if $n > 1$ then $n^6 + ... etc > 1$ so $(n+1) = p^2$ and $n^6 + ... etc = 1$ is out of the question.
That just leaves $n + 1 = n^6 + .... = p$ as an option. And for that to work we must have $n = 1$. And in that case it does work out that $n+1 = n^6 + ... etc = 2$.
answered Dec 2 '18 at 21:23
fleabloodfleablood
68.7k22685
$endgroup$
add a comment |
$begingroup$
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors.
Or fewer.
To rephrase and elaborate on greedoid's answer in a way that doesn't involve complete factoring:
So if if the polynomial factor as $r(n)q(n)$ we must have either $r(n) = 1; q(n) = p^2$ for some $n$ and prime $p$ or $r(n) =q(n)=p$ for some $n$ and prime $p$.
Note that if $n =-1$ we get the polynomial is $0$ so $n+1$ factors.
And $frac {n^7 + n^6 + n^5 + 1}{n+1} = n^6 + n^4 -n^3 +n^2 -n + 1$.
If $n+1 = 1$ then we get $n = 0$ and .... the other is $1$ and that won't do.
It's pretty clear if $n > 1$ then $n^6 + ... etc > 1$ so $(n+1) = p^2$ and $n^6 + ... etc = 1$ is out of the question.
That just leaves $n + 1 = n^6 + .... = p$ as an option. And for that to work we must have $n = 1$. And in that case it does work out that $n+1 = n^6 + ... etc = 2$.
answered Dec 2 '18 at 21:23
fleabloodfleablood
68.7k22685
$endgroup$
What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors.
Or fewer.
To rephrase and elaborate on greedoid's answer in a way that doesn't involve complete factoring:
So if if the polynomial factor as $r(n)q(n)$ we must have either $r(n) = 1; q(n) = p^2$ for some $n$ and prime $p$ or $r(n) =q(n)=p$ for some $n$ and prime $p$.
Note that if $n =-1$ we get the polynomial is $0$ so $n+1$ factors.
And $frac {n^7 + n^6 + n^5 + 1}{n+1} = n^6 + n^4 -n^3 +n^2 -n + 1$.
If $n+1 = 1$ then we get $n = 0$ and .... the other is $1$ and that won't do.
It's pretty clear if $n > 1$ then $n^6 + ... etc > 1$ so $(n+1) = p^2$ and $n^6 + ... etc = 1$ is out of the question.
That just leaves $n + 1 = n^6 + .... = p$ as an option. And for that to work we must have $n = 1$. And in that case it does work out that $n+1 = n^6 + ... etc = 2$.
answered Dec 2 '18 at 21:23
fleabloodfleablood
68.7k22685
answered Dec 2 '18 at 21:23
fleabloodfleablood
68.7k22685
answered Dec 2 '18 at 21:23
fleabloodfleablood
68.7k22685
answered Dec 2 '18 at 21:23
fleabloodfleablood
68.7k22685
68.7k22685
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