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Source: Indian Statistical Institute (ISI) UGB 2023 P7

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#1 h May 14, 2023, 1:19 PM • 4 Y

Y by SatisfiedMagma, GeoKing, Amir Hossein, Sarla

(a) Let $n \geq 1$ be an integer. Prove that $X^n+Y^n+Z^n$ can be written as a polynomial with integer coefficients in the variables $\alpha=X+Y+Z$ , $\beta= XY+YZ+ZX$ and $\gamma = XYZ$ .
(b) Let $G_n=x^n \sin(nA)+y^n \sin(nB)+z^n \sin(nC)$ , where $x,y,z, A,B,C$ are real numbers such that $A+B+C$ is an integral multiple of $\pi$ . Using (a) or otherwise show that if $G_1=G_2=0$ , then $G_n=0$ for all positive integers $n$ .

This post has been edited 1 time. Last edited by proxima1681, May 14, 2023, 1:40 PM
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#2 h May 14, 2023, 1:53 PM

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lifeismathematics

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lifeismathematics

#3 h Jul 7, 2023, 6:24 AM • 1 Y

Y by maths_arka

a)we proceed through proof by induction.

Base Case:( $n=1,2,3$ )

$(n=1)$
$X+Y+Z=\alpha=1\cdot \alpha +0\cdot \beta+0\cdot \gamma$ , so it is a polynomial with integer coefficients in variable $\alpha,\beta,\gamma$
$(n=2)$

$X^2+Y^2+Z^2=\alpha^2-2\beta+0\cdot \gamma$ which is a polynomial with integer coefficients in variable $\alpha,\beta,\gamma.$
$(n=3)$

$X^3+Y^3+Z^3=\alpha^3-3\alpha\cdot \beta+3\gamma$ which is a polynomial with integer coefficients in variable $\alpha, \beta, \gamma$ .

Now set $f_{n}(\alpha,\beta,\gamma)=X^{n}+Y^{n}+Z^{n}$ , for all positive integers $n$

We assume that given statement holds till case of $n-1$ i.e $f_{n-1}(\alpha,\beta,\gamma)$ is a polynomial with integer coeffiicients and variables $\alpha,\beta , \gamma$ .

Then we notice that $(X^{n-1}+Y^{n-1}+Z^{n-1})(X+Y+Z)=X^{n}+Y^{n}+Z^{n}+(XY+YZ+ZX)(X^{n-2}+Y^{n-2}+Z^{n-2})-XYZ(X^{n-3}+Y^{n-3}+Z^{n-3})$

i.e. $f_{n}(\alpha,\beta,\gamma)=\alpha\cdot f_{n-1}(\alpha,\beta,\gamma)-\beta\cdot f_{n-2}(\alpha,\beta,\gamma)+\gamma \cdot f_{n-3}(\alpha,\beta,\gamma)$

so from induction hypothesis step we have $f_{n}(\alpha,\beta,\gamma)$ to be a polynomial with integer coefficients in variables $\alpha,\beta , \gamma$ . $\blacksquare$

b) Consider $(xe^{iA})^{n}=x^{n}(\cos(nA)+i\sin(nA))$ so we have $G_{n}=\Im((xe^{iA})^{n}+(ye^{iB})^{n}+(ze^{inC})^{n})$

Consider $X=xe^{iA}, Y=ye^{iB}, Z=ze^{iC}$

so we have $G_{n}=\Im(X^{n}+Y^{n}+Z^{n})$

Also $G_{1}=G_{2}=0 \implies \Im(X+Y+Z)=0$ and $\Im(X^2+Y^2+Z^2)=0 \implies \Im(XY+YZ+ZX)=0$

and also $\Im(XYZ)=\Im(xyz e^{i(m\pi)})$ where $m \in \mathbb{Z}^{+}$ , so we have $\Im(XYZ)=0$

so from (a) we can write $X^{n}+Y^{n}+Z^{n}$ as polynomial with integer coefficinets with variable $X+Y+Z,XY+YZ+ZX, XYZ$ , so we have $\Im(X^{n}+Y^{n}+Z^{n})$ to be $0$ as Imaginary part of all variables is $0$ . so this implies $G_{n} \equiv 0$ for all positive integers $n$ . $\blacksquare$

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#5 h Jul 7, 2023, 9:20 AM

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Let $P_k(\alpha,\beta,\gamma)$ denotes the polynomial of $X^k+Y^k+Z^k$ with integer coefficients in the variable $(\alpha,\beta,\gamma)$ where $k\in \mathbb{N}$ .
We proceed by induction on n.
First we check that it holds for n=1, since $P_1(\alpha,\beta,\gamma)=\alpha$ and it has integer coefficients.
For n=2 we have $P_2(\alpha,\beta,\gamma)={\alpha}^2-2\beta$ and it also has integer coefficients. For n=3 we have $P_3(\alpha,\beta,\gamma)={\alpha}^3-3\alpha\beta+3\gamma$ and it also has integer coefficients.
Now its easy to see that $P_k(\alpha,\beta,\gamma)$ follows a recurrence relation as follows-
$P_{k+1}(\alpha,\beta,\gamma)=\alpha P_k(\alpha,\beta,\gamma)-\beta P_{k-1}(\alpha,\beta,\gamma)+\gamma P_{k-2}(\alpha,\beta,\gamma).$ So by strong form of induction we are done.
For second part we take
$X=xz_1$ , $z_1=\cos(A)+i\sin(A)$
$Y=yz_2$ , $z_2=\cos(B)+i\sin(B)$
$Z=zz_3$ , $z_3=\cos(C)+i\sin(C)$

where $(i^2=-1)$ .Here -
$\alpha=xz_1+yz_2+zz_3$ $\beta=(xz_1)(yz_2)+(xz_1)(zz_3)+(zz_3)(yz_2)$ $\gamma=xyz(z_1z_2z_3)$ We have
$X^n+Y^n+Z^n={(xz_1)}^n+{(yz_2)}^n+{(zz_3)}^n$
Now by De Moivre's Theorem we get
$X^n+Y^n+Z^n=x^n(\cos(nA)+i\sin(nA))+y^n(\cos(nB)+i\sin(nB))+z^n(\cos(nC)+i\sin(nC))$ Now according to the question we have to prove the imaginary part of $(X^n+Y^n+Z^n)=0 \forall n\geq 1$ or in other words $(X^n+Y^n+Z^n) \in\mathbb{R}$ $\forall n\geq 1$ .
Again we would proceed by induction.
For n=1 we have $G_1=0$ which means $(X+Y+Z)\in\mathbb{R}$ i.e $\alpha\in\mathbb{R}$ .
Now we also have $G_2=0$ which means
$(X^2+Y^2+Z^2)\in\mathbb{R}$ . Using the relation $(X^2+Y^2+Z^2)={\alpha}^2-2\beta$ we conclude that $\beta\in\mathbb{R}$ .
Now since we have $A+B+C$ is an integral multiple of $\pi$ we also have $\gamma \in\mathbb{R}$
(Just write $z=e^{i\theta}$ and check).
Using the relation $(X^3+Y^3+Z^3)={\alpha}^3-3\alpha\beta+3\gamma$ we conclude that $(X^3+Y^3+Z^3)\in\mathbb{R}$ .
Now using the recurrence we have got in part(1) and using strong form of induction we conclude that $(X^n+Y^n+Z^n)\in\mathbb{R}\hspace{0.2 cm}\forall n\geq 1$ .And hence $G_n=0 \hspace{0.2 cm}\forall n\geq 1$ .

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