Polynomial roots

Question

Let $f=nX^n+X^2-nX-1$. Show that $f$ has no roots in $\mathbb{Q}-\mathbb{Z},\:\forall n\geq5$. So, I was trying to consider $\alpha=\frac{p}{q}$ a root for $f$, with $(p,q)=1$ and trying to get a contradiction with our last condition, but I can not find anything. Is this the way to go for the problem or not? Thx a lot!

Answer 1

First note that

\begin{align} f(x)&=nx^n+x^2-nx-1\\ &=(x-1)(nx^{n-1}+nx^{n-2}+\cdots+nx+n)+x^2-nx-1+n\\ &=(x-1)(nx^{n-1}+nx^{n-2}+\cdots+nx+n)+(1-n+x)(x-1)\\ &=(x-1)(nx^{n-1}+nx^{n-2}+\cdots+nx^2+(n+1)x+1)\\ \end{align}

So we're looking for solutions to $$nx^{n-1}+nx^{n-2}+\cdots+nx^2+(n+1)x+1=0$$ or, after dividing by $n$, $$x^{n-1}+x^{n-2}+\cdots+x^2+\tfrac{n+1}nx+\tfrac 1n=0$$ Now let $x=\frac pq$, with $\gcd(p,q)=1$. Then: $$(\tfrac pq)^{n-1}+(\tfrac pq)^{n-2}+\cdots+(\tfrac pq)^2+\tfrac{n+1}n(\tfrac pq)+\tfrac 1n=0$$ which we may multiply by $q^{n-1}$ to obtain $$p^{n-1}+qp^{n-2}+\cdots+q^{n-3}p^2+\tfrac{n+1}nq^{n-2}p+\tfrac 1nq^{n-1}=0$$ Now note that

$$\tfrac{n+1}nq^{n-2}p+\tfrac 1nq^{n-1}=q^{n-2}p+\tfrac 1nq^{n-2}(p+q)$$

Now note that

$$p^{n-1}+qp^{n-2}+\cdots+q^{n-3}p^2+q^{n-2}p+\tfrac 1nq^{n-2}(p+q)=0$$

can be written as $$p(p^{n-2}+qp^{n-3}+\cdots+q^{n-3}p+q^{n-2})+\tfrac 1nq^{n-2}(p+q)=0$$ it is now clear that $\tfrac 1nq^{n-2}(p+q)$ is an integer and divisible by $p$. We may multiply that by $n$ to see $p$ divides $q^{n-2}(p+q)$; but this is a contradiction. Since $p$ and $q$ are coprime, $p$ has no factors in common with $q^{n-2}$, leading to $p\mid p+q$ which would mean $p\mid q$.

This may already seem like we're done; however, if $p=1$, we get $\gcd(p,q)=1$ for free, and $p\mid q$ suddenly isn't a contradiction anymore. Handling that separately, substituting $p=1$ yields

$$1+q+\cdots+q^{n-3}+q^{n-2}+\tfrac 1nq^{n-2}(1+q)=0$$

from this, we see that if we have a prime $r\mid q$ that doesn't divide $n$ (so $r$ divides the last term) we have a contradiction, since then all terms except the first are divisible by $r$, hence can never add up to be $0$. Thus, all primes that divide $q$ must divide $n$; this means $\gcd(n,q+1)=1$ so that since $\tfrac1nq^{n-2}(q+1)$ is an integer, we know $n\mid q^{n-2}$. Even stronger, since that prime $r\mid q$ cannot divide the last term, all primes in $q^{n-2}$ must be divided out by $n$ so that $\tfrac1nq^{n-2}=\pm1$. Now we can try to solve $q^{n-2}=\pm n$ (and if $q^{n-2}=n$ doesn't have solutions with $q>0$, neither does $q^{n-2}=\pm n$ for general $q$, so we'll do that one instead). First note that $q\neq \pm1$ (otherwise the fraction $\tfrac pq$ was an integer) so we may assume $q\geq 2$. Then we can proceed to prove $q^{n-2}> n$ for $n\geq 5$ using induction. I will leave that up to you.

Having done that, it means there are no rational solutions to $f(x)=0$ except $x=1$ whenever $n\geq 5$.