# If $N$ is large enough, then $x^5-Nx+1$ and $x^5-Nx^2+1$ are irreducible over $\mathbb{Q}$.

Show that if $$N$$ is large enough, then $$x^5-Nx+1$$ and $$x^5-Nx^2+1$$ are irreducible over $$\mathbb{Q}$$.

There is a hint for $$x^5-Nx+1$$: prove first that four of the roots in $$\Bbb{C}$$ has absolute values larger than $$1$$. I think that it follows from Rouche's theorem, but I don't know how to proceed.

Update: I think that both problems can be solved by direct computations. For example for the second one, we can write $$x^5-Nx^2+1=(x^3+ax^2+bx+u)(x^2+cx+v)$$where $$|u|=|v|=1$$ and $$a,b,c\in \Bbb{Z}$$. Multiply out, consider the coefficient of $$x^4$$ we get $$a+c=0$$. Consider $$x$$ we get $$uc=-bv$$. Next consider $$x^3$$ we have $$b+v+ac=0$$. Note that $$c=-a$$ and $$|b|=|a|$$, so we can actually solve $$a$$. Then a contradiction is easily reached.

I think that the same method also applies to the first one. However, the first one can be solved by using the fact that only one root is in the unit circle. I am still wondering if there is a nicer method for the second one.

Update 2: I got a solution from my professor. A similar method can also be applied to $$x^5-Nx^2+1$$. By Rouche's theorem, there are exactly two roots inside the unit circle, and the key point here is these two roots are complex conjugates. Hence they have to be the two roots of $$x^2+cx+v$$. Then the other three roots for $$x^3+ax^2+bx+u$$ are absolutely greater than one. The contradiction follows by Vieta's theorem. (See rtybase's answer below)

• Check this source, Proposition 6, for the number of routes. Commented Feb 29, 2020 at 11:38
• Otherwise it suffices to prove that $x^5-zx+1\in \Bbb{Q}(z)[x]$ is irreducible to obtain that its reduction $\bmod z-N$ is irreducible for all but finitely many $N$ (let $\Bbb{Q}(z,\alpha(z))$ be the splitting field then $x^5-Nx+1$ is reducible translates to $\sigma(\alpha(N))=\alpha(N)$ for $\sigma$ in one of the possible decomposition group and this is an algebraic equation in $N$). If $x^5-zx+1$ is reducible then every quintic is solvable contradicting that we know some quintic $\in \Bbb{Q}[x]$ with Galois group $A_5$. Commented Feb 29, 2020 at 12:55

For $$P(x)=x^5-Nx+1$$ ...

Gauss' lemma states that if a polynomial is irreducible over integers, it is also irreducible over rationals. Let's stick with integers.

Proposition 1. $$P(x)=(x-\alpha_1)(x-\alpha_2)...(x-\alpha_5)$$ has $$4$$ roots with $$|\alpha_i|>1$$.

Explained in the comments above, usig this source, Proposition 6. By the way, this part considers the "$$N$$ is large enough" criteria.

Proposition 2. $$P(x)$$ is irreducible over integers.

Let's suppose it's reducible, i.e. $$P(x)=G(x)\cdot F(x)$$, where $$G, F$$ are non-constant polynomials with integer coefficients. Then $$P(0)=G(0)\cdot F(0)=1$$. This means that the absolute values of the last coefficients of $$G$$ and $$F$$ are 1. Let's notes $$G(x)=g_kx^k+...+g_0, g_i\in \mathbb{Z}, g_k \ne 0, |g_0|=1$$ $$F(x)=f_mx^m+...+f_0, f_i\in \mathbb{Z}, f_m \ne 0, |f_0|=1$$ where $$k+m=5$$, $$k\geq1$$ and $$m\geq1$$

From Vieta's formulas we have that

• the absolute value of the product of $$G(x)$$'s roots ($$G(\beta_j)=0, \beta_j \in \mathbb{C}$$) is $$\left|\prod_{j=1}^k \beta_j \right|= \left|\frac{g_0}{g_k}\right|=\frac{1}{\left|g_k\right|}\leq 1 \tag{1}$$
• the absolute value of the product of $$F(x)$$'s roots ($$F(\gamma_t)=0, \gamma_t \in \mathbb{C}$$) is $$\left|\prod_{t=1}^m \gamma_t \right|= \left|\frac{f_0}{f_m}\right|=\frac{1}{\left|f_m\right|}\leq 1 \tag{2}$$

But $$\{\alpha_i\}_{i=1}^5=\{\beta_j\}_{j=1}^k \bigcup\{\gamma_t\}_{t=1}^m$$ and (from Proposition 1) only one root has $$|\alpha_{i_0}|\leq1$$ which falls either in $$\{\beta_j\}_{j=1}^k$$ or $$\{\gamma_t\}_{t=1}^m$$ (it can not belong to both, because $$P(X)=G(X)\cdot F(X)$$). But then:

• if $$\alpha_{i_0}\in \{\beta_j\}_{j=1}^k$$ then $$\prod\limits_{t=1}^m |\gamma_t | > 1$$ contradicting $$(2)$$
• if $$\alpha_{i_0}\in \{\gamma_t\}_{t=1}^m$$ then $$\prod\limits_{j=1}^k |\beta_t | > 1$$ contradicting $$(1)$$
• Thank you. It seems that the same method does not apply to $x^5-Nx^2+1$. Can I have a hint for the second question? Commented Feb 29, 2020 at 23:13
• @EricSun although, this problem is really interesting, I don't think I will have time to look for a neater solution for the 2nd problem. But, I am glad you have a better understanding of the problem and what needs to be done, more or less. Commented Mar 2, 2020 at 19:01

One can also see it by straight forward computation:

1. Conclude that $$p$$ cannot have a linear factor, thus if $$p$$ factors then $$p(x) =(x^3+ax^2+bx\pm1)(x^2-ax\pm1)$$ with $$a$$, $$b$$ in $$\mathbb Z$$.

2. Multiply out that representation of $$p$$ and compare coefficients at $$x^3$$ ... $$x$$ (at $$x^4$$ is already 0) against the required shape of $$p$$ to constrain $$N$$.

3. Conclude that the constraint from 2. cannot be satisfied by large $$N$$.