# Is the polynomial $g(t) = \sum_{q \text{ prime }, q\le p} t^{q-2}$ for a prime $p\ge 7$ separable? [closed]

Let $$a_0 + a_1 x^1 + a_2 x^2 + \cdots + a_n x^n$$ be a polynomial in $$\mathbb{Z}[x]$$ such that $$a_0 \neq 0$$ and $$|a_i| \le 1$$ for $$i=0,1,\cdots,n$$. Then my conjecture is that this polynomial is separable. Is there any way to prove this, or is this wrong in general? Thanks for your help.

Edit Is there any reason why $$g(t) = \sum_{q \text{ prime }, q\le p} t^{q-2}$$ for a prime $$p\ge 7$$ should be separable?

I can prove that any root has absolute value $$\le 2$$, there exists exactly one negative real root, zero positive real roots and all other complex roots come in conjugate pairs, since the degree is odd.

It seems that those polynomials (except for $$p=13$$) are also irreducible. I tried to apply Newton polygons to show irreducibility from which separability would follow, but without success.

Any help is highly appreciated.

## closed as off-topic by YCor, user44191, Sean Lawton, orgesleka, Dima PasechnikMar 31 at 16:52

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• Was there any basis for making your conjecture, e.g., had you looked at numerical data? I generated random degree 5 polynomials and hit a counterexample on the second try: $(x-1)(x^2+x+1)^2 = x^5 + x^4 + x^3 - x^2 - x - 1$. Four iterations later I found a second counterexample: $(x-1)^2(x+1)(x^2+x+1) = x^5 - x^3 - x^2 + 1$. It took a bit longer to find a counterexample in degree 6, but here is one: $(x-1)^2(x^4+2x^3+2x^2+x+1) = x^6 - x^4 - x^3 + x^2 - x + 1$. – KConrad Mar 31 at 8:50
• There is also a counterexample in degree $3$. $$(x-1)(x+1)^2=(x^2-1)(x+1)=x^3+x^2-x-1.$$ – Uriya First Mar 31 at 8:52
• @KConrad: I was looking at different polynomials which seem to also be (except one case I believe) irreducible. I thought that the more general statement would also be true. – orgesleka Mar 31 at 10:15

Actually with certain assumptions this will work. For example, Perron's theorem says that if $$a_n = 1$$ and $$\mid a_{n - 1} \mid > 1 + \mid a_{n - 2} \mid + \cdots + \mid a_{0} \mid$$ then the polynomial will be irreducible and hence separable. So it will be required for modulus of $$a_{n - 1}$$ to be at least $$2$$ for this theorem to work.
• As a TeX matter, you want to use \lvert\rvert, like $\lvert x\rvert > 1$ (\lvert x\rvert > 1), not \mid\mid, like $\mid x\mid > 1$ (\mid x\mid > 1); notice the spacing of the right bar relative to the inequality sign. – LSpice May 24 at 20:09