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I suspect that for some real nonnegative values $a_1, \dots, a_n$ the equation $$(x+1-a_1)(x+1-a_2)\cdots (x+1-a_n)=(-1)^n$$ has no roots that could be algebraically expressed using $a_1,\dots,a_n,n$.

I've found at least one set of such $a_i$'s, but they are complex; for $a^T=(1, 2, 0, 1+i, 1-i)$ we get: $(x+1-1)(x+1-2)(x+1-0)(x+1-1-i)(x+1-1+i)=(-1)^5=-1$

$x(x-1)(x+1)(x-i)(x+i)=-1$

$x(x^2-1)(x^2+1)=-1$

$x(x^4-1)=-1$

$x^5-x+1=0$

Wikipedia says this particular polynomial is an example of a quintic whose roots cannot be expressed in terms of radicals. This is what I'm looking for, however some of my $a_i$'s weren't real nonnegative numbers.

Where should I be looking for some sample $a_i$ withing the boundaries of my interest that would lead to a polynomial that is known to have no radical roots or other candidates than $x^5-x+1=0$ that would suit my needs? The Course in computational algebraic number theory by H. Cohen contains some mysterious list of polynomials in the Computing Galois Groups section but I was unable to make any sense of it.

By the way, if there is any nicer way of showing this, I would also highly appreciate it. Or maybe it is possible to solve the equation for all $\mathbb{R}_+$ numbers?

Thanks a lot!

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Any irreducible quintic that has exactly two nonreal zeros has Galois group $S_5$ and therefore its zeros can't be expressed in radicals. It's easy to make up examples of such polynomials of the forms $ax^5+bx^i+c$ for $i=1,2,3,4$ just using a bit of Calculus to force the two nonreal zeros, and Eisenstein or reduction modulo 2 to force the irreducibility. Whether any of the polynomials of these forms will satisfy your factorization property, I don't know, but it seems worth a try.

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  • $\begingroup$ Thanks, I'll investigate some more in this direction $\endgroup$ Oct 17 '16 at 18:19
  • $\begingroup$ Am I correct, that if we consider quintics of coefficients within $Q(i)={a+bi: a,b\in Q}$ then $(x-\sqrt{2}i)(x+\sqrt{2}i)(x-1)(x-2)(x-3)$ is irreducible (because root $\sqrt{2}$ is not in $Q(i)$, but solvable, since all roots are radicals of $Q(i)$? How should I understand the first sentence of your answer then? $\endgroup$ Oct 19 '16 at 11:55

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