How prove $P(x)=\sum_{k=0}^{n}(2k+1)x^k$ is irreducible over $\mathbb{Q}$ Show that the polynomial
$$P(x)=\sum_{k=0}^{n}(2k+1)x^k,\forall n\in N^{+}$$ is irreducible over $Q$.
My try: Since $P(x)$ has integer coefficients and the gcd of these coefficients is $1$, by Gauss's lemma it suffices to show that $P(x)$ is irreducible in $Z[x]$, and y this I can prove any complex zero of the polynomial satisfies $|z|<1$.
Because if $|z|\ge 1$, then since $z$ is a zero of $(1-z)P(z)$,we get
$$1+2z+2z^2+\cdots+2z^n-(2n+1)z^n=0.$$
Thus
$$|(2n+1)z^n|=|1+2z+2z^2+\cdots+2z^n|\le 1+2|z|+2|z^2|+\cdots+2|z^n|<|z^n|+2|z|^n+\cdots+2|z|^n=(2n+1)|z|^n$$
contradiction. Therefore $|z|<1$.
Then suppose that 
$$P(x)=f(x)g(x)$$
where $f$ and $g$ are nonconstant integer polynomials, then
$$1=P(0)=f(0)g(0)$$
then
$|f(0)|$ ,$|g(0)|$ equals $1$,
Say $|g(0)|=1$, let $b$ be the leading coefficient of $g$,if $\alpha_{1},\alpha_{2},\cdots,\alpha_{k}$  are the roots of $g$, then
$$|\alpha_{1}\alpha_{2}\cdots\alpha_{k}|=\dfrac{1}{|b|}\le 1$$
But  follow I can't find contradiction. So I can't prove this problem,
I find somebook, I can't find this similar problem, Thank you for you help
 A: This is not a full answer. I am going to use the following lemma due to Enestrôm and Kayeka( Proof, p.5 )

Lemma. Let $P=a_nX^n+\cdots+a_1X+a_0 \in \mathbb{R}[X]$ such that $a_i>0$ for all $i \in \{0,\cdots, n \} $. Then the roots of $P$ are contained in the annulus 
  $$
\min_{0\leq i< n}\left(\frac{a_i}{a_{i+1}}\right) \leq |z| \leq \max_{0\leq i< n}\left(\frac{a_i}{a_{i+1}}\right).
$$

Therefore by the lemma the root of $P(x)=\sum_{k=0}^{n}(2k+1)x^k$ are contained in the annulus 
$$
\frac{1}{3} \leqslant |z| < 1\tag{1}
$$

Let $P \in \mathbb{Z}[X] \setminus \{ \pm 1\}$ and let $m \in \mathbb{Z}$ such that $|P(m)|$ is 1 or prime number and such that $P$ has no roots in the disc $\{z\in \mathbb{C}:|z-m| \leqslant 1 \}$. Then $P$ is irreducible over $\mathbb{Z}$$\tag{2}$

Proof.
Suppose that $P=QR$ with $Q,R\in \mathbb{Z}[X]\setminus \{ \pm 1\}$. By assumption, we have
$|Q(m)|=1$ or $|R(m)|=1$. WLOG, suppose that $|Q(m)|=1$. Since $Q\neq \pm1$, $Q$ is not constant and we have write $Q=a\prod_{i=1}^r (X-\alpha_i)$ with $a\in \mathbb{Z}\setminus \{ 0\}$, $r\in\mathbb{N}$ and $\alpha_1,\cdots,\alpha_r$ are the roots of $Q$ counted by their multiplicity. Hence we have
$$
1=|Q(m)|=|a|\prod_{i=1}^r |m-\alpha_i|
$$
Since $a$ is a non-zero integer, we have $|a|\geq 1$, so that there exist $j\in\{1,\cdots, r \}$ such that $|m-\alpha_i|\leq 1$ and $P(\alpha_i)=Q(\alpha_i)R(\alpha_i)=0$, which is impossible since $P$ has no roots in the disc $\{z\in \mathbb{C}:|z-m| \leqslant 1 \}$. The proof is complete.


*

*Now take  $m=2$ and $(1)$. Thus $P$ has no roots in the disc $|z-2| \leqslant 1$. If $$P(2) = 2^{n+1}(2n-1)+3$$ is a prime number then $P$ is irreducible over $\mathbb{Z}$



Edit: An improvement

If $2n+1$ is a prime number then $P = (2n+1)X^n + \dotsb + 3X + 1$ is irreducible over $\mathbb{Z}$

Proof.
Let 
$$
Q(X)=X^n P(1/X) = X^n + 3X^{n-1} + \dotsb + (2n+1)
$$
Then all root are in the annulus  $|z| > 1$, now using (2) for $m=0$ the result is obvious.
Then, by Dirichlet's theorem we know there exist infinitely many primes of the form $2n+1$ 
A: Mine is just a partial answer, too.
As stated in the comments above, if $n+1$ is a prime and the order of $2 \pmod{n+1}$ is equal to $n$, then $P_n(x)$ is irreducible over $\mathbb{Q}$ since it is irreducible over $\mathbb{F}_2$. However, this rules out just a few cases, like $2n+1=p$ does. 
In a completely different framework, an interesting idea would be to show that $P_n(x)$ has an "isolated" zero inside the annulus $\frac{1}{3}\leq |z|<1$, in the following sense: there exists $a,b,c,d\in\mathbb{Z}$ such that all the zeroes of the meromorphic function $Q_n(z)=P_n\left(\frac{az+b}{cz+d}\right)$, except one, lie in $\{z:|z|<1\}$, then apply an adapted version of the lemma: 

If the monic polynomial $q\in\mathbb{Z}[x]$ has all its complex roots
  except one in $\{z:|z|<1\}$ and $|q(0)|$ is $1$ or a prime number,
  then $q(x)$ is irreducible over $\mathbb{Q}$

to $Q_n$ instead of $P_n$. Another idea is to consider to apply Eisenstein criterion to the numerator of
$$P_n\left(\frac{1+x}{1-x}\right)$$
whose coefficients appear to be very well-behaved mod the smallest prime dividing $n+1$. 
In facts, by defining $Q_n(x)=(1-x)^n\cdot P_n\left(\frac{1+x}{1-x}\right)$, we have:
$$Q_n(x)=\frac{(1-x)^{n+1}-(1+x)^{n+1}+2x(n+1)(1+x)^{n+1}}{2x^2},$$
and it is quite easy to check that if $n+1$ is a prime the only coefficient of $Q_n(x)$ that is not divisible by $(n+1)$ is the coefficient of $x^{n-1}$: this immediately gives that $P_n(x)$ is irreducible not only when $(2n+1)$ is a prime, but also when $(n+1)$ is a prime.
Also consider that, if $n$ is even, $Q_n(x)$ is just $(x+1)^n$ in $\mathbb{F}_2$; it should not be difficult to deduce the irreducibility of $P_n(x)$ also in this case. If $n$ is odd but not of the form $2^k-1$, the argument of the previous point should work by taking $p$ as the largest prime divisor of $n+1$, while if $n=2^k-1$ then $Q_n(x)$ is just $2^k$ times the polynomial $(1+x)^{2^k-1}$ in $\mathbb{F}_2$.
A: How about using reduction criterion? $f$ primitive in $\mathbb{Z}[t]$, with leading coefficient $a_n$ not divisible by a prime $p$. Consider the ring homomorphism $\mathbb{Z}[t] \to (\mathbb{Z}/p\mathbb{Z})[t] = F$ ($\bar{f}$ is the image of f.) If $\bar{f}$ is irreducible in $F[t]$, then $f$ is irreducible in $\mathbb{Q}[t]$.
