Việt's polynomial

First we'll show that if $y = x + \frac 1 x$ , then $a_n := x^n + \frac 1 {x^n}$ can be expressed as a polynomial in y of degree n, for n=1,2,3, ... .

For that, let $T^2 - aT + b = 0$ be the quadratic equation with roots x and 1/x. Then $b = x\cdot \frac 1 x = 1$ and $a = y$ . Since the quadratic equation has roots x and 1/x, we have:

$x^2 - yx + 1 = 0, \quad (1/x)^2 - y(1/x) + 1 = 0$ .

Multiplying the first equation by $x^n$ and the second by $x^{-n}$ , we get:

$x^{n+2} - yx^{n+1} + x^n = 0, \quad (1/x)^{n+2} - y(1/x)^{n+1} + (1/x)^n = 0$ .

Adding them together gives us $a_{n+2} = y\cdot a_{n+1} - a_n = y\cdot a_{n+1} - a_n$ . Starting from $a_0 = 2, a_1 = y$ , we get:

$a_2 = y^2 - 2, a_3 = y^3 - 3y, \ldots, a_7 = y(y^6 - 7y^4 +14y^2 -7).$

Hence, for $x = \sqrt[7]{\frac 3 5}$ , we have $\alpha = y$ and

$\alpha(\alpha^6 - 7\alpha^4 + 14\alpha^2 - 7) = \frac 3 5 + \frac 5 3 = \frac{34}{15}.$

This gives a rational polynomial of degree 7 for which $\alpha$ is a root. To show that this is the minimal polynomial over $\mathbb{Q}$ , we'll use some field theory. First, if $\beta = \sqrt[7]{3/5}$ , then $\mathbb{Q}[\beta]$ is an extension field of degree 7 over $\mathbb{Q}$ . But $\beta$ satisfies a quadratic equation over $\mathbb{Q}[\alpha]$ , via

$\alpha = \beta + \frac 1 \beta \implies \beta^2 - \alpha\beta + 1 = 0$ .

Hence

$7 = [\mathbb{Q}[\beta]:\mathbb{Q}] = [\mathbb{Q}[\beta] : \mathbb{Q}[\alpha]] \times [\mathbb{Q}[\alpha] : \mathbb{Q}]$ ,

where $[\mathbb{Q}[\beta] : \mathbb{Q}[\alpha]]$ is 1 or 2. This gives $[\mathbb{Q}[\alpha] : \mathbb{Q}] = 7$ as desired so the above degree-7 polynomial is indeed the minimal polynomial over $\mathbb{Q}$ . Hence:

$f(y) = -15 y(y^6 - 7y^4 + 14y^2 - 7)+34$

which gives f(1) = 19.

Because $α$ is a root of the $7th$ degree polynomial $f(x),$

$x$ $=$ $\sqrt[7]{\frac {3}{5}}$ $+$ $\sqrt[7]{\frac {5}{3}}$

From this equation we can get,

$x^2$ $=$ $\sqrt[7]{(\frac {3}{5})^2}$ $+$ $2$ $+$ $\sqrt[7]{(\frac {5}{3})^2}$

$x^2$ $-$ $2$ $=$ $\sqrt[7]{(\frac {3}{5})^2}$ $+$ $\sqrt[7]{(\frac {5}{3})^2}$

$(x^2$ $-$ $2)^2$ $=$ $\sqrt[7]{(\frac {3}{5})^4}$ $+$ $2$ $+$ $\sqrt[7]{(\frac {5}{3})^4}$

$(x^2$ $-$ $2)^2$ $-$ $2$ $=$ $\sqrt[7]{(\frac {3}{5})^4}$ $+$ $\sqrt[7]{(\frac {5}{3})^4}$ $\dots$ $(1)$

Also,

$x^3$ $=$ $\sqrt[7]{(\frac {3}{5})^3}$ $+$ $3\sqrt[7]{\frac {3}{5}}$ $+$ $3\sqrt[7]{\frac {5}{3}}$ $+$ $\sqrt[7]{(\frac {5}{3})^3}$

$x^3$ $-$ $3x$ $=$ $\sqrt[7]{(\frac {3}{5})^3}$ $+$ $\sqrt[7]{(\frac {5}{3})^3}$ $\dots$ $(2)$

From $(1)$ $\times$ $(2)$ we can get,

$((x^2 - 2)^2 - 2)$ $(x^3 - 3x)$ $=$ $(\sqrt[7]{(\frac {3}{5})^4}$ $+$ $\sqrt[7]{(\frac {5}{3})^4})$ $(\sqrt[7]{(\frac {3}{5})^3}$ $+$ $\sqrt[7]{(\frac {5}{3})^3})$

$((x^2 - 2)^2 - 2)$ $(x^3 - 3x)$ $=$ $\frac {3}{5}$ $+$ $\sqrt[7]{\frac {3}{5}}$ $+$ $\sqrt[7]{\frac {5}{3}}$ $+$ $\frac {5}{3}$

$((x^2 - 2)^2 - 2)$ $(x^3 - 3x)$ $=$ $x$ $+$ $\frac {34}{15}$

$-15$ $((x^2 - 2)^2 - 2)$ $(x^3 - 3x)$ $+$ $15x$ $+$ $34$ $=$ $0$

Hence, $f(x)$ $=$ $-15$ $((x^2 - 2)^2 - 2)$ $(x^3 - 3x)$ $+$ $15x$ $+$ $34$

Substitute $x$ $=$ $1,$ we get $f(1)$ $=$ $19$

Suppose that x= $(p)^{1/7}$ + $(1/p)^{1/7}$ , where p= 3/5 So:

$x^2$ -2 = $(p)^ {2/7}$ + $(1/p)^{2/7}$ ...(1)

$x^3$ - $3x$ = $(p)^{3/7}$ + $(1/p)^{3/7}$ ...(2)

$x^4$ - $4x$ +2 = $(p)^{4/7}$ + $(1/p)^{4/7}$ ....(3)

$x^5$ - $5x^{3}$ + $5x$ = $(p)^{5/7}$ + $(1/p)^{5/7}$ ....(4)

Then take a power of 7 in the both sides to obtain:

$x^7$ = $((p)^{1/7}$ + $(1/p)^{1/7}$ ) $^7$

Expand the right side to get:

$x^7$ = $(7(1/p)^{1/7}$ $p^{6/7}$ + $21(1/p)^{2/7}$ $p^{5/7}$ + $35*(1/p)^{3/7}$ $p^{4/7}$ + $35*(1/p)^{4/7}$ $p^{3/7}$ + $21*(1/p)^{5/7}$ $p^{2/7}$ +p+ $7*(1/p)^{6/7}$ $p^{1/7}$ + $1/p$

$x^7$ = p+(1/p)+ $7p^{5/7}$ + $(7/p)^{5/7}$ + $21p^{3/7}$ + $(21/p)^{3/7}$ + $(35p)^{1/7}$ + $(35/p)^{1/7}$

Because p=3/5, so p+1/p=34/15, then multiply both sides with 15 to get:

$15x^7$ =34+15( $7(p^{5/7}$ + $(1/p)^{5/7})$ )+ $15*21$ ( $(p)^{3/7}$ + $(1/p)^{3/7}$ )+ $15*35$ ( $p^{1/7}$ + $(1/p)^{1/7}$ )

Substituting all the equations (1),(2),(4) to obtain:

$15x^7$ =34+105( $x^5$ - $5x^3$ + $5x$ )+ $15*21$ ( $x^3$ -3x)+ $15*35x$

f(x)= $-15x^7$ + $105(x^5-5x^3+5x)$ + $15*21(x^3-3x)$ + $15*35x$ +34

Thus, substituting x=1 to get:

f(1)= -15+105+ $15*21$ + $15*35$ +34

f(1)= -15+34

f(1)= 19

We will proceed directly by finding $f(x)$ . Since $\alpha$ is a root of this polynomial, it must be possible to express $\alpha^7$ as a sum of monomials $\alpha^k$ of lower degree with rational coefficients. Let us write $b = \sqrt[7]{3/5}$ so that $\alpha = b + 1/b$ and compute

$\alpha^7 = b^7 + 7 b^5 + 21 b^3 + 35 b + 35 / b + 21 / b^3 + 7 / b^5 + 1/b^7 .$

Our strategy now will be to succesively subtract different powers of $\alpha$ so that only the rational term $b^7$ remains. So compute further

$\alpha^5 = b^5 + 5 b^3 + 10 b + 10 / b + 5 / b^3 + 1 / b^5$

and

$\alpha^3 = b^3 + 3 b + 3 / b + 1 / b^3 .$

It is then not hard to see that

$\alpha^7 - 7 \alpha^5 + 14 \alpha^3 + 7 \alpha = b^7 + 1/b^7 = 3/5 + 5/3 .$

But this means that we have found the polynomial with integer coefficients that $\alpha$ is a root of, namely

$f(x) = -15( x^7 - 7x^5 + 14 x^3 + 7x -34/15).$

Now we simply plug in $x=1$ to get $f(1) = -15 + 34 = 19$ .

The Binomial Expansion gives:

• $(x+\frac {1}{x})^{7}=x^{7}+7(x^{5}+\frac {1}{x^{5}})+21(x^{3}+\frac {1}{x^{3}})+35(x+\frac {1}{x})$
• $(x+\frac {1}{x})^{5}=(x^{5}+\frac {1}{x^{5}})+5(x^{3}+\frac {1}{x^{3}})+10(x+\frac {1}{x})$
• $(x+\frac {1}{x})^{3}=(x^{3}+\frac {1}{x^{3}})+3(x+\frac {1}{x})$

Combining these give:

• $(x+\frac {1}{x})^{7}=x^{7}+\frac {1}{x^{7}}+7(x+\frac {1}{x})^{5}-14(x+\frac {1}{x})^{3}+7(x+\frac {1}{x})$

Putting $x=\sqrt [7]{\frac {5}{3}}$ gives:

• $\alpha^{7}=\frac {34}{15}+7\alpha^{5}-14\alpha^{3}+7\alpha$

Hence Viet's polynomial is:

• $f(x)=34+105x^{5}-210x^{3}+105x-15x^{7}$

So $f(1)=19$

There is a question of uniqueness, but it is possible to show that f is irreducible.

Let $p=\sqrt[7]{3/5}$ . We then have $\sqrt[7]{5/3} = 1/p$ and we can write $\alpha = p + 1/p$ . We now want to find the minimal polynomial for $\alpha$ . We can make an educated guess that the degree of this polynomial will be 7.

Taking the seventh power of the above equation gives us the following:

$\alpha^7 = p^7 + 7p^5 + 21p^3 + 35p + 35\left(\frac{1}{p}\right) + 21\left(\frac{1}{p}\right)^3 + 7\left(\frac{1}{p}\right)^5 + \left(\frac{1}{p}\right)^7$

Parts of this look nice: $p^7 = 3/5$ , $\left(\frac{1}{p}\right)^7=(5/3)$ , and $35p + 35\left(\frac{1}{p}\right) = 35\alpha$ . We still have to deal with the fifth and the third powers of $p$ and $1/p$ .

In order to do that, let's take the fifth power of the original equation:

$\alpha^5 = p^5 + 5p^3 + 10p + 10\left(\frac{1}{p}\right) + 5\left(\frac{1}{p}\right)^3 + \left(\frac{1}{p}\right)^5$

Scaling it and subtracting from the equation for the seventh power, we get:

$\alpha^7 - 7\alpha^5 = p^7 -14 p^3 - 35p - 35\left(\frac{1}{p}\right) - 14\left(\frac{1}{p}\right)^3 + \left(\frac{1}{p}\right)^7$

And now we do the same with the equation for $\alpha^3$ to obtain the following:

$\alpha^7 - 7\alpha^5 + 14\alpha^3 = p^7 + 7p + 7\left(\frac{1}{p}\right) + \left(\frac{1}{p}\right)^7$

And that can now be rewritten as

$\alpha^7 - 7\alpha^5 + 14\alpha^3 = (3/5) + 7\alpha + (5/3)$

From this expression and the condition that the leading coefficient of the polynomial should be -15 we see that $\alpha$ is the root of the following polynomial:

$p(x) = -15x^7 + 105x^5 - 210x^3 + 105x + 34$

Thus, $p(1) = -15+105-210+105+34 = 19$ .

For convenience, let $z=x+\dfrac{1}{x}$ . Note that we have $x^3+\dfrac{1}{x^3} = \left(x+\dfrac{1}{x}\right)^3 - 3x-\dfrac{3}{x} = z^3-3z$ , and that $x^5 + \dfrac{1}{x^5} = \left(x+\dfrac{1}{x}\right)^5 -5x^3-10x-\dfrac{10}{x} - \dfrac{5}{x^3}$ $=z^5 - 5(z^3-3z) - 10z = z^5-5z^3+5z$ , obtained by expanding the expressions in the parentheses via Binomial Theorem.

Then we have that $x^7+\dfrac{1}{x^7} = \left(x+\dfrac{1}{x}\right)^7 -7x^5-21x^3-35x-\dfrac{35}{x}-\dfrac{21}{x^3} - \dfrac{7}{x^5}$ $=z^7 - 7(z^5-5z^3+5z) - 21(z^3-3z) - 35z = z^7-7z^5+14z^3-7z$ (*).

If we assign $x=\sqrt[7]{\dfrac{3}{5}}, \dfrac{1}{x} = \sqrt[7]{\dfrac{5}{3}}$ , then clearly $x+\dfrac{1}{x} = \alpha$ , and also $x^7+\dfrac{1}{x^7} = \dfrac{3}{5}+\dfrac{5}{3} = \dfrac{34}{15}$ . Additionally, according to the equation (*), since $z=\alpha$ we have that $x^7+\dfrac{1}{x^7} = \dfrac{34}{15} = \alpha^7-7\alpha^5+14\alpha^3-7\alpha$ . Moving everything over to one side and multiplying by $-15$ yields the equation $-15\alpha^7 + 105\alpha^5 - 210\alpha^3 + 105\alpha+34=0$ , and so $\alpha$ is a root of this 7th degree polynomial, which satisfies the conditions of the problem, as it has integer coefficients and a leading coefficient of $-15$ . Thus, we have found $f(x)$ . Substituting $x=1$ yields $f(1) = -15+105-210+105+34 = \boxed{19}$ .

First we will find a polynomial with integer coefficients of degree $7$ , with leading coefficient $-15$ and with $\alpha = \sqrt[7]{\frac{3}{5}}+\sqrt[7]{\frac{5}{3}}$ as a root. To fully solve the problem, it must be shown that this polynomial is unique, or at least that its value at $1$ is unique. This will be discussed briefly at the end.

Let $y = \sqrt[7]{\frac{5}{3}}$ so that $\alpha = y + 1/y$ . Since we are after a polynomial of degree $7$ , we will look at the value $\alpha^7$ . This is

$(y+1/y)^7 = (y^7 + y^{-7}) + 7(y^5 + y^{-5}) + 21(y^3 + y^{-3}) + 35(y + y^{-1}).$

On the right hand side, there are two things we know already. The value $y + 1/y$ is simply $\alpha$ and $y^7+y^{-7}$ is just $3/5+5/3 = 34/15$ . Let's examine the remaining terms $y^3+y^{-3}$ and $y^5+y^{-5}$ . We see that

$(y+1/y)^3 = (y^3 + y^{-3}) + 3(y+y^{-1})$

so

$y^3+y^{-3} = \alpha^3 - 3\alpha.$

Similarly,

$(y+1/y)^5 = (y^5+y^{-5}) + 5(y^3+y^{-3}) + 10(y+y^{-1}).$

So

$y^5+y^{-5} =\alpha^5 - 5(y^3+y^{-3}) + 10\alpha$ $= \alpha^5 - 5(\alpha^3-3\alpha) + 10\alpha$ $= \alpha^5- 5\alpha\%3 - 5\alpha.$

Plugging in these values for $y^3+y^{-3}$ and $y^5+y^{-5}$ into our earlier equation for $\alpha^7$ we find that

$\alpha^7 = 7\alpha^5 -14 \alpha^3 + 7 \alpha +34/15$

and so

$f(x) = -15x^7 + 7x^5 - 14x^3 + 7x + 34$

has $\alpha$ as a root, and all of the requirements as phrased in the problem. Furthermore we have

$\boxed{f(1) = 19.}$

Now returning to our point of the uniqueness of the polynomial $f(x)$ with the prescribed properties. To do this we will assume some knowledge of the theory of field extensions. The element $\alpha$ lies in the field $\alpha \in \mathbb Q(\sqrt[7]{3},\sqrt[7]{5})$ , the field of rationals $\mathbb Q$ extended to contain both the $7$ th root of $3$ and $5$ . This field extension has degree $49 = 7^2$ over $\mathbb Q$ and any element of this field satisfies some irreducible polynomial with rational coefficients of some degree dividing the degree of the extension, $49$ . We have shown that $\alpha$ satisfies a polynomial of degree $7$ so either this is irreducible, or $\alpha$ satisfies a polynomial of degree $1$ over $\mathbb Q$ , i.e. $\alpha \in \mathbb Q$ . If $\alpha = y + 1/y \in \mathbb Q$ then $y$ satisfies the quadratic polynomial with rational coefficients $y^2 - \alpha y + 1 = 0$ , but by the same reasoning, $y$ is the root of either an irreducible polynomial of degree $7$ or $1$ and it clearly $y \not \in \mathbb Q$ , so $y$ cannot satisfy a polynomial of degree $2$ over $\mathbb Q$ , and so we must have $\alpha \not \in \mathbb Q$ .

first we move from leading coefficient let $\sqrt[7]{\frac {5}{3}}= a$ and $\sqrt[7]{\frac {3}{5}}=b$ so $\alpha=a+b$ and we get $ab=1$ let $f(x)=-15x^7$ and we know that $0=f(\alpha)=-15\alpha^7=-15(a+b)^7$ $0=-15(a^7+7ab(a^5+b^5)+21(ab)^2(a^3+b^3)+35(ab)^3(a+b)+b^7$ so we get $0=-15(a^7+7(a^5+b^5)+21(a^3+b^3)+35(a+b)+b^7)$ $0=-34-105(a^5+b^5)-315(a^3+b^3)-525(a+b)$ we must erase $-105(a^5+b^5)$ in left side so we must add $105x^5$ in $f(x)$ so we get $f(x)=-15x^7+105x^5$ if we use binomial newton we can get $0=-34+210(a^3+b^3)+525(a+b)$ we must erase $210(a^3+b^3)$ in left side so we must add $-210x^3)$ in $f(x)$ so we get $f(x)=-15x^7+105x^5-210x^3$
so we can get $0=-34-105(a+b)$ equation same if we add $34+105x$ in $f(x)$ so we get $f(x)=-15x^7+105x^5-210x^3+105x+34$ $f(1)=-15(1)^7+105(1)^5-210(1)^3+105(1)+34$ $f(1)=-15+105-210+105+34=19$

For this problem, we want to find the sum of the coefficients of f(x) . First, consider what happens with the 7th order term when we plug in the root. By the binomial theorem, we have $-15(\sqrt[7]{\frac{3}{5}} + \sqrt[7]{\frac{5}{3}})^7 = -(9+105(\frac{3}{5})^{5/7} + \cdots + 25)$ Since we have only integer coefficients, the only term in f(x) that can cancel the -(9+25) = -34 is the constant term. This makes the sum of known coefficients so far -15+34=19. I believe we should be able to argue from symmetry at this point that the middle coefficients must cancel. However, the rigorous argument eludes me, so we shall proceed to find the rest of the polynomial with some algebraic computation. This is more work, but it gives us infinitely more information than the problem asked for, which is pretty neat. The next highest term in the polynomial must be $105x^5$ in order to cancel $(\frac{3}{5})^{5/7}$ that appeared above. Expanding $105(\sqrt[7]{\frac{3}{5}} + \sqrt[7]{\frac{5}{3}})^5$ and seeking to cancel terms, we get that the next term must be -210x^3. This is a bit cumbersome to write out, so I will stick to this sketch. One more application of binomial theorem and collecting terms finishes the problem to give $f(x) = -15x^7 + 105x^5 - 210x^3 + 105x +34.$ Thus, $f(1) = 19.$ I happened to note that $f(2) =4$ , which is quite small. The root that we began with must be close to 2.

Let x=\alpha^{1/7}+\frac{1}{\alpha^{1/7}} where \alpha = \frac{3}{5}

The following identities can be established:

\alpha^{2/7}+\frac{1}{\alpha^{2/7}} = x^{2}-2 \alpha^{3/7}+\frac{1}{\alpha^{3/7}} = x^{3}-3x \alpha^{4/7}+\frac{1}{\alpha^{4/7}} = x^{4}-4x+2 \alpha^{5/7}+\frac{1}{\alpha^{5/7}} = x^{5}-5x^{3}+5x

Coupled with the binomial expansion for x^{7} = (\alpha^{1/7}+\frac{1}{\alpha^{1/7}})^{7} we get:

f(x) = -15(x^{7}-7x^{5}+14x^{3}-7x)+34 So that f(1)=19