# Is there a way to determine if the n-th roots of a polynomial is a polynomial?

I was this problem: $$\int\frac{dx}{\sqrt{x^4+2x^3+3x^2+2x+1}}$$

I solved this question because I just knew that $$(1+x+x^2)^2=x^4+2x^3+3x^2+2x+1$$ but this made me wonder is there is a way to know if the $$n-$$th root of a certain polynomial is a polynomial? Given a polynomial how to determine $$n$$ where the $$n-th$$ root of this polynomial is a polynomial and how to determine the $$n-$$th root of this polynomial?

There is Taylor theorem or the extended binomial theorem that can find $$(P(x))^{1/n}$$ but this is not what I am looking for because

1. It only deals with two terms and the more terms that a polynomial have the more one needs to use the binomial theorem on each term which gets very ugly.
2. It gives an infinite series which doesn't tell if each the $$n-$$th root is a polynomial or not
3. Even if one can make simplifications on the infinite series to conclude it is a polynomial finding suitable $$n$$ would need one to check all the cases from $$2-m$$ where $$m$$ is the degree of polynomial, which is not very effective.
• I would plug in some small numbers of $x$. In your example, $P(x)= x^4+2x^3+3x^2+2x+1$ we have $P(-2)= 9, P(-1)= 1, P(0)=1, P(1)=9$ so I suspect maybe it is a square of other polynomial. Commented May 25 at 7:45
• There is a long-division-like algorithm for calcluating the square root of a number. This can be converted into an algorithm for calculating the square root of a polynomial, in a similar way that long division itself has been converted into an algorithm for polynomial division. Commented May 25 at 8:03
• For perfect squares see How do we check if a polynomial is a perfect square? Some algorithms for perfect powers are reviewed in Giesbrecht-Roche, Detecting lacunary perfect powers and computing their roots. Commented May 25 at 8:32
• The discriminant is zero if the polynomial has a repeated root; so if it is nonzero then the polynomial won't be a perfect power. Commented May 25 at 8:38
• There is a multinomial theorem to address your point 1. I'm not sure it really helps, though. Commented May 25 at 8:44

If $$f(x)$$ is a power of something, then $$f(x)$$ and $$f'(x)$$ have that something as a common divisor. In your case, $$\gcd(x^4 + 2 x^3 + 3 x^2 + 2x + 1, 4x^3+6x^2+6x+2) \\ = x^2 + x + 1 \text{.}$$

More generally, $$f(x) = (p(x))^n$$ implies $$f'(x) = n(p(x))^{n-1}p'(x)$$, so $$(p(x))^{n-1}$$ divides both $$f$$ and $$f'$$, therefore divides their greatest common divisor.

Note that this also applies to powers of non-polynomials. $$\frac{\mathrm{d}}{\mathrm{d}x} \sin^4(x) = 4 \sin^3(x)\cos(x)$$ and $$\sin(x)$$ divides both of these. It can be tricky to define gcds for non-polynomials. Your experience with trigonometric identities should suggest that the "immediately visible" common factors may appear, vary, and disappear as we manipulate the expressions.

If everything in sight is a polynomial, then the leading term of $$f$$ must be an $$n^{\text{th}}$$ power and so must the constant coefficient. For your example $$f$$, the leading term requires $$n \in \{1,2,4\}$$ and the constant term doesn't restrict $$n$$. The fact that the gcd in your example is not a cube and is not a constant tells us $$n = 2$$. Though the constant coefficient doesn't restrict $$n$$, if $$f$$ has a polynomial $$n^{\text{th}}$$ root, the constant tells us that root's constant coefficient is an $$n^{\text{th}}$$ root of $$1$$.

Let's see if we can cobble this into an algorithm. (Not optimal. Not efficient. Probably not even particularly good. But at least a procedure that works.) We will discuss starting with a polynomial having, integer ($$\Bbb{Z}$$), rational ($$\Bbb{Q}$$), real ($$\Bbb{R}$$) or complex ($$\Bbb{C}$$) coefficients and target $$n^{\text{th}}$$ roots from the same list. The indeterminate will be $$x$$ and the collections of polynomials will be denoted $$\Bbb{Z}[x]$$, $$\Bbb{Q}[x]$$, $$\Bbb{R}[x]$$, and $$\Bbb{C}[x]$$, respectively. We assume we have a polynomial $$f(x)$$ from one of these collections and we search for an $$n \in \Bbb{Z}$$ with $$n \geq 1$$ and a polynomial $$p(x)$$ from one of the collections such that $$f(x) = (p(x))^n$$.

For this discussion:

• The phrase "$$f(x)$$ is a first power" means that $$n$$ can only be $$1$$ and $$p(x)$$ must be $$f(x)$$.
• $$A$$ stands for "either $$\Bbb{Z}$$ and $$\Bbb{Q}$$", since almost always these are treated together. Similarly, $$A[x]$$ is used for the collection of polynomials with integer or rational coefficients. (By Gauss's lemma, these are really equivalent factorizations possibly wearing two different hats.)
• $$B[x]$$ stands for "either $$\Bbb{R}$$ or $$\Bbb{C}$$" since almost always these are treated together. Similarly, $$B[x]$$ is used for this collection of polynomials.
• Whenever we write a polynomial as $$\text{[coefficient]}x^{\text{[power]}} + \cdots$$, the elided terms have lower degree than the power shown. That is, we have written the leading term and suppressed the non-leading terms.

Let's start by maybe making $$f$$ a little simpler. Then continue by inspecting the leading and trailing coefficients of $$f$$. We finish off by applying calculus a few times.

1. If the leading term of $$f$$ is negative, factor a $$-1$$ from $$f$$ and proceed below with $$-f$$. If we eventually find $$-f = p^n$$ and $$n$$ is odd then $$f = (-p)^n$$; otherwise, $$f$$ is a first power.

2. Inspect the leading term. If searching for a $$p(x) \in A[x]$$, factor the coefficient. $$n$$ must be a divisor of each power of a prime and the power of $$x$$. If searching for $$p(x) \in B[x]$$, $$n$$ must be a divisor of the power of $$x$$. (The coefficient does not constrain because in $$B$$, everything has whatever root(s) we might desire.)

Examples:

• $$f(x) = 32x^{10} + \cdots$$: The leading term is $$2^5 x^{10}$$. If we want $$p(x) \in A[x]$$, then $$n \in \{1,5\} \cap \{1,2,5,10\} = \{1,5\}$$, so $$f$$ is a first or fifth power. If we want $$p(x) \in B[x]$$, then $$n \in \{1,2,5,10\}$$.
• $$f(x) = \frac{32}{81}x^{10} + \cdots$$: The leading term is $$\frac{2^5}{3^4} x^{10}$$. If we want $$p(x) \in A[x]$$, $$n \in \{1,5\} \cap \{1, 2, 4\} \cap \{1,2,5,10\} = \{1\}$$, so $$f$$ is a first power. If we want $$p(x) \in B[x]$$, $$n \in \{1,2,5,10\}$$.
• $$f(x) = \sqrt{2} x^{11} + \cdots$$: The leading coefficient is not in $$A$$, so $$p(x) \not \in A[x]$$. So we search for $$p(x) \in B[x]$$ and have $$n \in \{1,11\}$$.
3. If we seek $$p(x) \in A[x]$$, then also factor the constant coefficient, with sign. For instance, if the constant coefficient is $$1$$, it has infinitely many factorizations: $$1^k$$ and $$(-1)^{2k}$$ for any $$k \geq 0$$ in $$\Bbb{Z}$$, leading to $$n \geq 1$$ in $$\Bbb{Z}$$. We always insert $$n = 1$$ by hand. Instead, if we seek $$p(x) \in B[x]$$, the constant coefficient does not constrain $$n$$; we have $$n \in \Bbb{Z}$$, $$n \geq 1$$.

Examples, where we take $$k$$ to range over positive integers and we seek $$p(x) \in A[x]$$:

• The constant coefficient is $$4 = (-1)^{2k} 2^2$$. Then $$n \in \left(2\Bbb{Z} \cap \{1,2\}\right) \cup\{1\} = \{1,2\}$$.
• The constant coefficient is $$-4 = (-1)^{2k+1} 2^2$$. Then $$n \in \left( (2\Bbb{Z} + 1) \cap \{1,2\} \right) \cup \{1\} = \{1\}$$.
4. Take the intersection of the candidate $$n$$s from the leading coefficient and the candidate $$n$$s from the constant coefficient (if the constant actually leads to a constraint). The set of surviving candidates form a lattice (of divisors of the largest candidate, ordered by "divides").

(The candidates from step 1 form a sublattice of the division lattice of the nonnegative integers. The candidates from step 2 do also. The intersection of two such lattices is another such lattice, so the set of survivors is such a sublattice.)

Let $$N$$ be the number of surviving candidates, sort the surviving candidates, and label them in increasing order $$n_1 < n_2 < \cdots < n_{N} \text{.}$$

Example, where $$60$$ is the largest candidate:

$$N = 12$$ and $$n_1 = 1$$, $$n_2 = 2$$, $$\dots$$, $$n_7 = 10$$, $$\dots$$, $$n_{N} = 60$$.

5. Suppose $$f = p^n$$ with $$n > 1$$ and degree of $$p$$ greater than one. Let's see what we get for the leading term when we differentiate. Let $$p(x) = a x^d + \cdots$$. Then \begin{align*} f(x) &= (p(x))^{n} \\ &= (a x^d + \cdots)^{n} \\ &= a^n x^{dn} + \cdots \text{ and } \\ f'(x) &= n (p(x))^{n-1} p'(x) \\ &= n (a x^d + \cdots)^{n-1} (ad x^{d-1} + \cdots) \\ &= n a^n x^{dn-1} + \cdots \text{.} \end{align*} So the leading term of $$f'$$ divided by the leading term of $$f$$ is $$n/x$$. If $$f$$ is not a power, then the leading term of $$f'$$ divided by the leading term of $$f$$ is still some constant over $$x$$, so again, this is a candidate $$n$$.

Either this candidate is on the list from step 3 or it is not. If not, $$f$$ is a first power and we are done. We've narrowed the list of possible powers to one.

6. Continuing the calculus from step 4 \begin{align*} f''(x) &= n(n-1)(p(x))^{n-2}p'(x)^2 + n (p(x))^{n-1}p''(x) \\ &= n(n-1)(ax^d + \cdots)^{n-2}(adx^{d-1}+\cdots)^2 + n(ax^d+\cdots)^{n-1}(ad(d-1)x^{d-2}+\cdots) \\ &= a^n dn(dn -1)x^{dn-2} + \cdots \end{align*} So the leading term of $$f''$$ divided by the leading term of $$f'$$ is \begin{align*} \frac{a^n dn(dn -1)x^{dn-2}}{n a^n x^{dn-1}} &= \frac{d(dn -1)}{x} \end{align*} This is, it's the degree of $$p$$ times one less than the degree of $$f$$ divided by $$x$$. Since we start with knowledge of the degree of $$f$$, we can divide the number in the numerator by one less than this degree and get $$d$$. Since $$d$$ has to be a positive integer, if the division has nonzero remainder, we discover $$f$$ is a first power and stop.

We can also check that the degree of $$d$$ times the one surviving candidate $$n$$ gives the degree of $$f$$. (I think this second check might be redundant given the prior work, but haven't thought about it hard enough to be sure.)

7. At this point, we have one candidate $$n$$ for $$f = p^n$$. Compute the polynomial GCD $$p(x) = \mathrm{gcd}\left( \left( \frac{\mathrm{d}}{\mathrm{d}x}\right)^{n-1} f(x), f(x) \right) \text{.}$$ If $$p$$ is a constant polynomial, $$f$$ is a first power. Otherwise, $$f = p^n$$. (And remember to apply the conclusion of step 0 if you factored a $$-1$$ out of $$f$$ in that step.)

(One might say that computing the $$(n-1)^{\text{th}}$$ derivative is a lot of work, but it can be done in one step if one has a table of falling factorials, since $$(\mathrm{d}/\mathrm{d}x)^{n-1} ax^b = a(b)_{n-1}x^{b-(n-1)}$$ if $$b \geq n$$ and zero otherwise.)

• This will detect powers, but it's worth noting that it will also detect some non-powers, so once we find the GCD, we should investigate further to see if it actually has the right form to tell us that $f(x)$ is a perfect power. For example, if we have $f(x)=x^4+2x^3-2x-1$, then $f'(x)=4x^3+6x^2-2$, then $\gcd(f(x),f'(x)) = x^2+2x+1=(x+1)^2$, but $f(x)$ is not a power of $(x+1)^2$; rather, it is $(x+1)^3(x-1)$. Commented May 25 at 20:11
• @MishaLavrov : I don't see that this differs from what I have written. Although maybe you point out something that wasn't sufficiently explicit? Commented May 26 at 5:10
• Yes - everything you wrote was correct and I'm not disagreeing with it, I just think this additional detail is worth mentioning. Commented May 26 at 15:39
• What it detects is repeated facrors of any sort. If there are nonrepeated factors or factors are repeated different numbers of times, you generally don't have a pure power. Commented May 27 at 10:42
• @MishaLavrov : Explicitness should be markedly increased. (And I foresee new comments about polishing the described procedure.) Commented Jun 2 at 15:06

No there isn't a general way as far as I know. But there is one trick that I would like to tell you.

The polynomial you have $$x^4+2x^3+3x^2+2x+1$$
is a symmetric polynomial so there is a possibility that it can be factorised. Let's check. You can use this trick for other symmetric polynomials.

Rewrite the whole expression as $$\displaystyle x^2\left(x^2+2x+3+\frac{2}{x}+\frac{1}{x^2}\right)$$

Now you just have to club the appropriate terms. Rewrite this as

$$\displaystyle x^2\left(\left(x+\frac{1}{x}\right)^2+2\left(x+\frac{1}{x}\right)+1\right)$$

This is nothing but $$\displaystyle x^2\left(x+\frac{1}{x}+1\right)^2$$

which certainly equals to $$(1+x+x^2)^2$$

• You seem to be implying that all symmetric polynomials are factorizable, but that's definitely not true: consider $x^2+bx+1$ for any $|b|<2$. Commented May 25 at 8:00
• Let me change that real quick Commented May 25 at 8:01
• You mean, "let me change that non-real quick." Commented May 25 at 18:30
• @PerAlexandersson It got changed, look at the edit history, now he says that "there is a possibility" Commented May 27 at 11:06

Suppose $$x^4+2x^3+3x^2+2x+1=(ax^2+bx+c)^2$$. Then expanding the right side and equating like-power terms renders

$$a^2=1$$

$$2ab=2$$

$$2ac+b^2=3$$

$$2bc=2$$

$$c^2=1$$

The first equation has two roots $$a\in\{1,-1\}$$ and wlog we may take either one. Selecting $$a=1$$ then leads to the second equation giving $$b=1$$ and the third equation giving $$c=1$$. If these results satisfy both remaining equations then you may render

$$x^4+2x^3+3x^2+2x+1=(x^2+x+1)^2.$$

A "real-life" example comes from Benjamin and Snyder's neusis construction of the regular hendecagon[1], which would be impossible with Euclidean tools plus an angle trisector alone. Their method ultimately depends on factoring a sextic polynomial

$$u^6+4u^5+8u^4+12u^3+12u^2+8u+4,$$

in which the degree is even, the constant term is a square and the coefficients add up to another square. To prove that this in fact is $$(ax^3+bx^2+cx+d)^2$$, we expand and match:

$$a^2=1\implies a=1$$ (wlog)

$$2ab=4\implies b=2$$

$$2ac+b^2=8\implies c=2$$

$$2(ad+bc)=12\implies d=2$$

$$2cb+c^2=12$$

$$2cd=8$$

$$d^2=4$$

where the values of $$a,b,c,d$$ from the first four equations have to match the last three — which actually works, and thus as the authors put it, "a miracle occurs".

Reference

1. Benjamin, Elliot; Snyder, C. Mathematical Proceedings of the Cambridge Philosophical Society 156.3 (May 2014): 409-424.; https://dx.doi.org/10.1017/S0305004113000753