polynomialsroots
Can a $5$-th grade polynomial have only one solution? for example:
$$x^5 – 3x^4 + 17x^3 – 12x^2 – 11x – 5 = 0$$
I mean that it's not necessary for every seventh grade polynomial to have seven solutions. There may be only one or three. The same for a sixth grade polynomial, there may be only two solutions.
If this is true, then how I can decide if a fifth grade polynomial has only one solution or three and not five solutions?
A polynomial is solvable iff its splitting field is solvable iff its Galois group is solvable. The Galois group of the cyclotomic polynomial $\Phi_n(x)$ is $(\mathbb{Z}/n\mathbb{Z})^{\ast}$, which is abelian, and all abelian groups are solvable.
An interesting question is how to find "nice" radical expressions for roots of unity. One way to do this is to go through the proof that the roots of a solvable polynomial are expressible in radicals by finding a composition series of the Galois group and constructing Kummer extensions.
This is perhaps easiest to describe by example, so take $n = 5$. Then $\Phi_5(x) = x^4 + x^3 + x^2 + x + 1$ has Galois group $(\mathbb{Z}/5\mathbb{Z})^{\ast} \cong C_4$, so it has a composition series with two factors of $C_2$. This implies that $\mathbb{Q}(\zeta_5)$ is a quadratic extension of a quadratic extension. Basic facts about Gauss sums imply that the second quadratic extension is $\mathbb{Q}(\sqrt{5})$, so $\zeta_5$ satisfies a quadratic polynomial with coefficients in $\mathbb{Q}(\sqrt{5})$. This polynomial must be $$x^2 - (\zeta_5 + \zeta_5^{-1}) x + 1.$$
Again, basic facts about Gauss sums imply that $\zeta_5 + \zeta_5^{-1} = \frac{-1 + \sqrt{5}}{2}$, so it follows by the quadratic formula that $$\zeta_5 = \frac{ 1 - \sqrt{5} + i \sqrt{10 + 2 \sqrt{5}} }{4}.$$
In general one may need to take more than just square roots, in which case things get complicated. There will be expressions one can write down generalizing Gauss sums that are guaranteed to land in a specific subfield of $\mathbb{Q}(\zeta_n)$ but I don't know a good way of actually figuring out what those expressions are by hand, and in writing down Kummer extensions one may need to adjoin smaller roots of unity (so I believe the most important case is when $n$ is prime).
A basic remark is that the problem reduces to the case that $n$ is a power of a prime, since for arbitrary $n$ it is possible to write $\zeta_n$ as a product of roots of the form $\zeta_{p^k}$ where $p^k \parallel n$.
After more thought, I feel sure enough to claim in an answer that $1^{1/5}$ is not a "solution in radicals" to $x^5-1=0$ in the sense that is guaranteed to exist by the solvability of the equation's Galois group. I think the kind of solution that the solvability of the Galois group implies exists is exactly a "non-trivial" solution in your sense.
The "solution in radicals" that is guaranteed by the solvability of the Galois group of a polynomial $f$ is really a "root tower" over $\mathbb{Q}$: a tower of fields, beginning with $\mathbb{Q}$ and ending with a field containing $f$'s splitting field, in which each field is obtained from the last one by adjoining a $p$th root for some prime $p$.
$$\mathbb{Q}\subset\mathbb{Q}[r_1]\subset\dots\subset\mathbb{Q}[r_1,\dots,r_k]$$
where for each $r_j$ there is a prime $p_j$ such that $r_j^{p_j}$ was already in the previous field $\mathbb{Q}[r_1,\dots,r_{j-1}]$ but $r_j$ is not. The mechanism of the proof is that if the Galois group $G$ is solvable, then there is a composition series
$$G=G_0 \triangleright G_1\triangleright \dots \triangleright G_k=\{1\}$$
such that each factor group $G_{j-1}/G_j$ is cyclic of prime order $p_j$. By the fundamental theorem of Galois theory, there is thus a tower of fields
$$\mathbb{Q}=K_0 \subset K_1 \subset \dots \subset K_k=\mathrm{splittingfield}(f)$$
where $K_j$ has degree $p_j$ over $K_{j-1}$. By possibly adjoining some extra elements, it can be guaranteed that each field extension can be accomplished by adjoining a $p_j$th root to the previous field (in other words a root $r_j$ of the polynomial $x^{p_j}-a$ where $a$ is in the previous field), so that the tower has the above form (a "root tower"). Because extra elements may have been added, the final field may now be bigger than $f$'s splitting field. (I am sweeping some possibly pertinent details under the rug here: the extra elements you need to adjoin are precisely the $p_j$th roots of unity. The argument, when applied to cyclotomic polynomials, is saved from circularity by induction on the size of the primes.)
The key point is this: when $p_j$th roots are adjoined, the field extensions have degree $p_j$. This means that the polynomial $x^{p_j}-a$ of which $r_j$ is a root must be irreducible over $K_{j-1}$. If the polynomial were reducible, $r_j$ would be the root of a polynomial of degree lower than $p_j$ and $K_j=K_{j-1}[r_j]$ would have degree less than $p_j$ over $K_{j-1}$.
What I'm getting at is that the $p$th roots that are adjoined at each step in such a "solution by radicals" are all roots of polynomials $x^p-a$ that are irreducible over the previous field. In particular, $$ \mathbb{Q} \subset \mathbb{Q}[1^{1/5}]$$ is not a solution of $x^5-1$ by radicals in this sense, because $x^5-1$ isn't irreducible over $\mathbb{Q}$. In fact, in this context as you note, $1^{1/5}$ doesn't even have an unambiguous meaning. It could be $1$, in which case this is not even a nontrivial field extension. On the other hand it could be any of the four primitive fifth roots of unity, and then the expression $\mathbb{Q}[1^{1/5}]$ would have a meaning determined up to isomorphism, but it still wouldn't be a "solution by radicals" because fifth roots of unity do not solve an irreducible polynomial of the form $x^p-a$.
Furthermore, the solutions to an equation furnished by a "solution by radicals" of this kind are always "non-trivial" in your sense. Since at each stage the adjoined root $r_j$ is a root of an irreducible polynomial over $K_{j-1}$, replacing it with any other root of this polynomial induces an automorphism of $K_j$ fixing $K_{j-1}$ pointwise. At the end of everything, the roots of $f$ (the original polynomial to be solved) are elements of the top field $\mathbb{Q}[r_1,\dots,r_k]$; thus they are rational expressions in $r_1,\dots,r_k$ with rational coefficients. Replacing any of the $r_j$'s with a different root of the same irreducible polynomial it was a root of thus induces an automorphism of the top field $\mathbb{Q}[r_1,\dots,r_k]$ that fixes $\mathbb{Q}$ pointwise. Thus any expression in $r_1,\dots,r_k$ that solves $f$ will still solve $f$ after this replacement.
To make the connection to the question explicit, this means that the cyclotomic polynomials, because they have solvable Galois groups, do all have legitimate / "non-trivial" solutions in radicals, and this result dates back to the days of Gauss and Galois.
Best Answer
A 5th degree polynomial (with real coefficients) has at least 1 real root. A polynomial of odd degree has at least 1 real root.
The fundamental theorem of algebra says that a polynomial roots equal to its degree. However, they may be complex and they may be roots of multiplicity.
$x^2-1$ has 2 real roots. $x^2+1$ has 2 complex roots. $x^2-2x + 1$ has one root of multiplicity 2.
Does this help?