Show product of $(r^5-r+2)$ over the 5 roots of $x^5+x^3+1=0$ is $1$

Question

Let $(a,b,c,d,e)$ be the five roots of $x^5+x^3+1=0$, and let $g(x) = x^5-x+2$.

Prove $$ g(a)g(b)g(c)g(d)g(e) = 1 $$

preferably without pages of messy algebra.

Answer 1

We'll do some computations. As noted in G. Sassatelli's comment, for roots of our polynomial,

$$x^5-x+2 - (x^5+x^3+1)=-x^3-x+1.$$ Use polynomial division to get $$x^2+1 + x^2(x^3+x-1)=x^5+x^3+1=0$$ and $$-(x^3+x-1)=(1+x^2)/x^2,$$ since none of the roots are zero. Realize that $abcde=-1$, so our product is equivalent to $$h(a)\cdots h(e)$$ with $h(x)=(1+x^2)$.

Use polynomial division again:

$$(1+x^2)x^3+1=x^5+x^3+1=0$$

$$(1+x^2)=-1/x^3$$

So our product is equivalent to $$(-1)^5/(a^3b^3c^3d^3e^3)=1.$$

Please tell me if there is a sign error.

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I believe there is a sort of euclidean algorithm behind all this here that will allow you to compute $g(a)\cdots g(e)$ for any $g$.

Answer 2

Edit: I misused Bezout's identity. It applies to the polynomial field $Q[x]$, not the polynomial ring $Z[x]$ because it is not a PID. The following is valid if it happens $b(x) \in Z[x]$. Which it does in the OP's case.

Theorem: Suppose $f,g \in Z[x]$ and $\gcd(f,g)=1$. Then $$\prod_{f(\alpha)=0} g(\alpha) = \pm 1.$$

Proof: Because the gcd is 1, we have $$a(x)f(x)+b(x)g(x)=1.$$ Thus, if $f(\alpha)=0$, we have $b(\alpha)g(\alpha) =1$. And hence

$$\prod_{f(\alpha)=0} b(\alpha) \cdot \prod_{f(\alpha)=0} g(\alpha) = 1.$$

The product of $g(\alpha)$ must be integer since it is symmetric and can be written in terms of the coefficients of $f$. Likewise for the product of $b(\alpha)$. The above implies then that

$$\prod_{f(\alpha)=0} g(\alpha) = \pm 1.$$ QED.

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In this question's case, it is easy to show the gcd of the relevant polynomials is 1. What remains is to find the sign. I am thinking about this.

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Also realize, that if $f$ is irreducible, as long as $g$ is not a multiple of $f$, then $\gcd(f,g)=1$. Also realize that for $\gcd(f,g)=a(x)$ with $a(x)$ non-constant, the product should be zero. The last case, when $\gcd(f,g)=c$, means that the product should be a power of $c$.

Answer 3

By that previous question, the result is an integer. Which integer? Just compute it numerically with sufficient accuracy, and round to the nearest integer. I'll leave it to you to estimate how much accuracy is sufficient.

Alternatively, it's a very quick computation in a Computer Algebra system, e.g., Maple:

> simplify(product(r^5-r+2, r = RootOf(_Z^5+_Z^3+1)));

$$1$$

Answer 4

Partial solution:

In the number field $\mathbb{Q}(r)$, $-r^3-r+1$ is invertible,

$$(1+r^3)(-r^3-r+1)=1$$ thus

$N(-r^3-r+1)$ is a unit so $N(-r^3-r+1)=\pm 1$

I dont know how to find the sign.