Find all natural numbers $n$, such that polynomial $n^7+n^6+n^5+1$ would have exactly 3 divisors.

Question

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.

Answer 1

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.

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How I got this this factorization: \begin{eqnarray} (n^7+n^5)+(n^6+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}

Answer 2

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 $n+1 = n^6 + ... etc = 2$.