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I saw this problem on a website a while ago and I 'm still stuck.

Let $\alpha = \large\sqrt[7] \frac{3}{5}+\sqrt[7] \frac{5}{3}$.

Find and prove uniqueness of a polynomial $P \in \mathbb Z[X]$, with degree $7$ and leading coefficient $-15$ such that $P(\alpha)=0$

It definitely has something to do with properties of $r+\frac{1}{r}$.

Any hint is welcome.

Gabriel Romon
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  • Sorry if I am missing sth obvious but I am thinking that $P(x)=-15 x^7+15 {\alpha}^7$ is satisfying the requirements. So we must prove uniqueness of that polynomial? – Mangomath Apr 26 '14 at 20:13
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    @Mangomath: I doubt $16\alpha^7$ is an element of $\mathbb Z$. – Dustan Levenstein Apr 26 '14 at 20:15
  • Of course you are right... – Mangomath Apr 26 '14 at 20:16
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    Let $T_7$ be the Chebyshev polynomial of the first kind of degree $7$. Then $T_7(\alpha/2) = \left(\tfrac{3}{5}+\tfrac{5}{3}\right)/2$. That leads (after multiplication with a suitable constant) to a polynomial $P$ as required. – WimC Apr 26 '14 at 20:16
  • @WimC Hmmm I'm clueless how you get this result... I can't think of way to derive it from $T_7(cos(x))=cos(7x)$ – Gabriel Romon Apr 26 '14 at 20:21
  • http://www.wolframalpha.com/input/?i=minimum+polynomial+of+%283%2F5%29%5E%281%2F7%29+%2B+%285%2F3%29%5E%281%2F7%29 – David P Apr 26 '14 at 20:23
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    @GabrielR. A more useful property is $$T_n\left(\frac{x+x^{-1}}{2}\right)=\frac{x^n+x^{-n}}{2}.$$ – WimC Apr 26 '14 at 20:23
  • @WimC Ok, once you know this... Anyway, how do you prove it in an elegant way ? I know the expanded form of $T_n$ but computations just seem too ugly. – Gabriel Romon Apr 26 '14 at 20:26
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    @GabrielR. By the recursive definition of $T_n$ and induction: $$T_{n+1}\left(\frac{x+x^{-1}}{2}\right) = (x+x^{-1})\left(\frac{x^n+x^{-n}}{2}\right) - \left(\frac{x^{n-1}+x^{-n+1}}{2}\right) = \frac{x^{n+1}+x^{-n-1}}{2}$$ – WimC Apr 27 '14 at 06:46

1 Answers1

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We may use the identity $$\eqalign{ (X+Y)^7-X^7-Y^7&=7XY(X+Y)(X^2+XY+Y^2)^2\cr &=7XY(X+Y)((X+Y)^2-XY)^2 } $$ So choosing $X=\root{7}\of{3/5}$ and $Y=1/X$ so that $\alpha=X+Y$ we get $$ \alpha^7-\frac{3}{5}-\frac{5}{3}=7\alpha(\alpha^2-1)^2 $$ and this reduces to $$ 34+105\alpha-210\alpha^3+105\alpha^5 -15\alpha^7=0. $$ Now, let us prove that $$ P(X)=34+105X-210X^3+105X^5 -15 X^7 $$ is Irreducible in $\Bbb{Z}[X]$ (or equivalently in $\Bbb{Q}[X]$). Indeed, If $$Q(X)=X^7P(1/X)=34X^7+105X^6-210X^4+105X^2 -15$$ Then clearly $5$ divides all the coefficients of $Q$ except the leading one, and $25$ does not divide the constant term. This proves that $Q$ is irreducible in $\Bbb{Z}[X]$ according to Eisenstein Irreducibility criterion. So the same holds for $P$.

Now the fact that $P$ is irreducible $\Bbb{Z}[X]$, proves that it is the minimal polynomial of $\alpha$ (because otherwise the minimal polynomial would be a factor of $P(X)$), and this minimality implies the uniqueness statement of your question. $\qquad\square$

Omran Kouba
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