Explain the proof :

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Explain the proof :

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Help me understand the corollary
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consider if we have \(n+1\) values \(a_0,\dots,a_n\). if the functions match at all points, we have: $$P(a_0)=Q(a_0)\\P(a_1)=Q(a_1)\\\dots\\P(a_n)=Q(a_n)$$ for a particular \(P(a_i)=Q(a_i)\), we can equivalently write $$P(a_i)-Q(a_i)=0$$ define the polynomial function \(R(x)=P(x)-Q(x)\), so the earlier list of equations tells us:$$R(a_0)=0\\R(a_1)=0\\\dots\\R(a_n)=0$$ so it follows that \(R\) has \(n+1\) zeros. but since \(R(x)=P(x)-Q(x)\), it follows that \(R\) is at most dimension \(n\)

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Other answers:

at most degree* n
so it follows that we must have that \(R(x)=0\) everywhere (in other words, in the representation \(R(x)=c(x-x_1)\dots(x-x_n)\), we must have \(c=0\))
and if \(R(x)=0\), well \(R(x)=P(x)-Q(x)\) so that means \(P(x)-Q(x)=0\implies P(x)=Q(x)\)
Okay for the $$P(a_0)=Q(a_0)$$ what does it basically explain the value of the function being equal or the term $$a_0$$
it means the two different polynomials \(P,Q\) match at \(x=a_0\)
in the first part why is it taken that 6|K please explain :)
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I want to understand the approach in the question. Please help
\(\epsilon_{1,2}\) are roots of unity -- recall that: $$\cos(\pm\pi/3)=\pm\frac12\\\sin(\pm\pi/3)=\frac{\sqrt3}2\\$$
so we have: $$\epsilon_1=\cos(\pi/3)+i\sin(\pi/3)$$now by de Moivre's theorem we see $$\epsilon_1^3=\cos(3\cdot\pi/3)+i\sin(3\cdot \pi/3)=\cos(\pi)+i\sin(\pi)=-1$$ and so \(\epsilon_1^6=(\epsilon_1^3)^2=(-1)^2=1\)
similarly, $$\epsilon_2=\cos(-\pi/3)+i\sin(-\pi/3)\\\epsilon_2^3=\cos(-\pi)+i\sin(-\pi)=-1\\\epsilon_2^6=(\epsilon_2^3)^2=(-1)^2=1$$
equivalently, you could've just done de Moivre's theorem directly for the sixth power, so: $$\epsilon_1^6=\cos(6\cdot \pi/3)+i\sin(6\cdot \pi/3)=\cos(2\pi)+i\sin(2\pi)=1\\\epsilon_2^6=\cos(6\cdot-\pi/3)+i\sin(6\cdot-\pi/3)=\cos(-2\pi)+i\sin(-2\pi)=1$$
and yes @ParthKohli I meant to say: $$\cos\left(\pm\frac{\pi}3\right)=\frac12\\\sin\left(\pm\frac\pi3\right)=\pm\frac{\sqrt3}2$$
anyways, they factor \(x^2-x+1=(x-\epsilon_1)(x-\epsilon_2)\) because that way we can check if \(x^2-x+1\) divides \(x^n-x+1\) by merely checking if the individual factors \(x-\epsilon_1,x-\epsilon_2\) divide \(x^n-x+1\). by the factor theorem, we know that \(x-\epsilon\) is a factor of \(f(x)=x^n-x+1\) if and only if \(f(\epsilon)=0\); in other words, $$\epsilon^n-\epsilon+1=0$$ so in this case we're checking both factors \(x-\epsilon_1,x-\epsilon_2\) so we have: $$\epsilon_1^n-\epsilon_1+1=0\\\epsilon_2^n-\epsilon_2+1=0$$
anyways, since \(\epsilon_1,\epsilon_2\) are roots of \(x^2-x+1=0\), consider: $$x^2-x+1=0\\x(x-1)=-1\\x(1-x)=1$$so it follows that \(x(1-x)=1\) or \(1-x=\frac1x=x^{-1}\) for \(x\in\{\epsilon_1,\epsilon_2\}\) -- in other words, $$1-\epsilon_1=1/\epsilon_1\\1-\epsilon_2=1/\epsilon_2$$
also i had the \(n\)-th degree polynomial wrong, I meant to write \(f(x)=x^n+x-1\) so $$\epsilon_1^n+\epsilon_1-1=0\\\epsilon_2^n+\epsilon_2-1=0$$
in which case we can rewrite the equations like: $$\epsilon_1^n=1-\epsilon_1\\\implies \epsilon_1^n=1/\epsilon_1\\\epsilon_2^n=1-\epsilon_2\\\implies \epsilon_2^n=1/\epsilon_2$$
in other words, \(\epsilon_1^{n+1}=1\) and \(\epsilon_2^{n+1}=1\) by multiplying both sides by \(\epsilon_1,\epsilon_2\) respectively
and yet we saw that: $$\epsilon_1^6=\epsilon_2^6=1$$so taking both sides to any integer power \(i\) we have $$\epsilon_1^{6i}=\epsilon_2^{6i}=1$$ so it follows that we must have \(n+1=6i\implies n=6i-1\)
thank u :)
the reason I showed earlier that \(\epsilon_1^3=\epsilon_2^3=-1\) earlier is to show that \(k=6\) is the *smallest* positive integer such that \(\epsilon_1^k=\epsilon_2^k=1\), otherwise we could've had other solutions we missed
oops, I said show and earlier too many times
so that \(i\) in \(6i-1\) has nothing to do with the imaginary unit \(i\) haha!

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