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Consider following polynomials: \begin{align} P_n(x)&=x^{2n-1}-\sum_{i=0}^{n-1}2\binom{2n}{2i+1}x^{2i} \end{align} for $n\geq 1$. Can we prove they are irreducible over $\mathbb{Q}$?

So first couple are $$x-4$$ $$ x^3-8x^2-8$$ $$ x^5-12x^4-40x^2-12$$ $$ \dots$$

For small specific values of $n$ we can prove it, for example for $x^3-8x^2-8$ we only need to check for linear factor, specifically for roots $\pm 1,\pm2,\pm4\pm 8$. With Maple I've verified more than first $600$ values of $n$ so far, all appear to be irreducible (using the irreduc function).

However for general $n$ I have no idea how to go about this, the Eisenstein criterion cannot be directly used, so I've tried substitutions like $P_n(x\pm 1)$, but it does not help.

Edit: Just out of curiosity, I have tried to change the polynomial a bit to see when it stops to be irreducible (again for small values of $n$, let's say to $100$) and I found it interesting that you can do quite a lot changes and it still remains irreducible. For example replacing the multiple of $2$ by other integers, or changing signs of the individual terms in sum, or even replace $2i+1$ in binomial coefficients with $2i$. What is going on?

Sil
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2 Answers2

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For odd $n$ the irreducibility of these polynomials follows from a generalization of Eisenstein's criterion known as Newton's polygon applied to the prime $p=2$.

When $n$ is odd the constant term $4n$ is divisible by four but not divisible by eight. Furthermore, Lucas' theorem tells us that all the binomial coefficients $\binom {2n}{2i+1}$ are even, implying that all the other coefficients, save for the leading $1$, are also all divisible by four (and possibly by a higher power of two). These imply that the Newton's polygon consists of a single line segment from $(0,2)$ to $(2n-1,0)$. The slope of this line is $-2/(2n-1)$ implying that the line segment does not pass through any integer points other than its endpoints. By $2$-adic theory this guarantees that $P_n(x)$ is irreducible over the $2$-adic field $\Bbb{Q}_2$ and hence also over the subfield $\Bbb{Q}$.

The case of an even $n$ is more complicated but it may be possible study the divisibility of the coefficients by powers of two and again settle the claim. For example the Newton's polygon of $$ P_4(x)=x^7-16(x^6+7x^4+7x^2+1) $$ is the line segment from $(0,4)$ to $(7,0)$, has slope $-4/7$, and the same argument goes through.

Jyrki Lahtonen
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  • I suspect that Kummer's theorem comes in handy for the study of even $n$. Gotta go. Hopefully someone can take it from here. – Jyrki Lahtonen Apr 26 '18 at 18:53
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    Very nice, I will have to study more the $p$-adic theory to see why this guarantees the result, do you know of some good literature for this? Also if I understood correctly, the very first even case $P_2(x)=x^3-8x^2-8$ leads to line segment from $(0,3)$ to $(3,0)$ which passes through some integer points, maybe this might make this method difficult for solving generic even case... – Sil Apr 26 '18 at 19:22
  • I, too, noticed that this argument doesn't apply to $P_2$. I'd guess that there are other similar cases. – Jyrki Lahtonen Apr 26 '18 at 19:29
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    For example with $P_8$ we get a line segment from $(0,5)$ to $(15,0)$. This passes through $(3,4)$, $(6,3)$, $(9,2)$ and $(12,1)$ leaving irreducibility in doubt. – Jyrki Lahtonen Apr 26 '18 at 19:35
  • I have tried to find all even $n$'s for which this works, see my answer. Hopefully it is correct, although even then there are still infinitely many $n$'s to go... – Sil Apr 28 '18 at 23:09
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Based on @JyrkiLahtonen's answer which proves the irreducibility for odd $n$ and suggests path or even $n$, I was able to prove the statement for even $n$'s for which the Newton polygon's method works, although it still leaves infinity of even $n$'s to check (hopefully someone can bring another idea to solve these as well).

The Newton polygons method seems to work for even $n$'s for which $(2n-1,v_2(4n))=1$, so $ n=4,6,10,12,16,18,20,22,\dots $.

First, we need to show that the Newton's polygon for $P_n(x)$ is indeed a line segment between the end points. For this it is sufficient to show that $v_2(2\binom{2n}{2i+1}) \geq v_2(4n)$ for $i=0,1,\dots,n-1$ (in other words every $y$ coordinate in the Newton polygon will be at least as large as the $y$ coordinate corresponding to the constant coefficient $4n$) . This can be written as $v_2(\binom{2n}{2i+1}) \geq v_2(2n)$. But this follows almost immediately if we notice that $(2i+1)\binom{2n}{2i+1}=2n\binom{2n-1}{2i}$, and so by applying the $v_2$ on both sides (using that $2i+1$ is odd) we get $$ v_2\left(\binom{2n}{2i+1}\right) = v_2\left(2n\binom{2n-1}{2i}\right) \geq v_2(2n). $$

Finally because the constant coefficient is $-4n$ and leading coefficient is $1$, the end points of the Newton polygon are $(0,v_2(4n))$ and $(2n-1,0)$, and so the method will guarantee irreducibility for $(2n-1,v_2(4n))=1$.

Sil
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