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!
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.