Write a full solution,

Find all roots of \(x^{5}+x^{4}+x^{3}-x^{2}-x-1 = 0\).

Prove that \(\displaystyle \left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^{n} = \cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)\)

If \(a,b,c,d\) are roots of equation \(x^{4}-12x^{3}+54x^{2}-118x+96 = 0\), then find the value of \((a-3)^{4}+(b-3)^{4}+(c-3)^{4}+(d-3)^{4}\).

Let \(P(x),Q(x)\) be polynomial with real coefficients such that \(\deg(P(x)) > \deg(Q(x))\). Find all solutions \((P(x),Q(x))\) that satisfy \(P(x)^{2}+Q(x)^{2} = x^{6}+1\).

This note is a part of Thailand Math POSN 2nd round 2015

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TopNewestSolution for Q-3:Using binomial theorem, one can note that,

\[P(x)=(x-3)^4-10x+15\]

We have, using Remainder-Factor Theorem and the fact that \(a,b,c,d\) are roots of \(P(x)\) that,

\[P(x)=(x-3)^4-10x+15=0~\forall~x\in\{a,b,c,d\}\implies (x-3)^4=10x-15~\forall~x\in\{a,b,c,d\}\]

Using the last result, we have the sum as,

\[\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)\]

Using Vieta's formulas, we have,

\[\displaystyle \sum_{x\in\{a,b,c,d\}} (x) = 12\\ \implies \sum_{x\in\{a,b,c,d\}} (10x)=120\]

Hence, the required sum evaluates as,

\[\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)=\left\{\left(\sum_{x\in\{a,b,c,d\}} (10x)\right)-\left(\sum_{x\in\{a,b,c,d\}} (15)\right)\right\}=120-15\times 4=120-60=\boxed{60}\]

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Nicely done.

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Your hint did most of the work. So, the credit actually goes to you. :)

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\(Q-1\)

\(x^5+x^4+x^3-x^2-x-1 = (x-1)(x^4+2x^3+3x^2+2x+10)\)\(\implies \text{one root is 1}\)

\(x^4+2x^3+3x^2+2x+1=0\implies x^2+2x +3+\dfrac{2}{x}+\dfrac{1}{x^2}=0\) \(\implies \left(x+\dfrac{1}{x}\right)^2-2+2\left(x+\dfrac{1}{x}\right)+3=0\)

\(x+\dfrac{1}{x} = m\)

\(m^2+2m+1=0 \implies \left(m+1\right)^2=0\implies m=1,1\)

\(x+\dfrac{1}{x}=1 \text{On solving using quadratic equations we get x}=\omega,\omega^2\)

(\(\implies \text{we get 4 roots from here }\omega,\omega,\omega^2,\omega^2\))

roots are \(\implies\) \(\boxed{\omega,\omega^2,1,\omega,\omega^2}\)

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You can factor it the easier way: \(x^5 + x^4 + x^3 - x^2 - x - 1 = x^3(x^2 + x^2 + 1) - (x^2 + x + 1) = (x^3-1)(x^2+x+1) \)

\( \Rightarrow (x-1)(x^2+x+1)^2 = 0 \Rightarrow x = 1, \omega, \omega^2 \)

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What motivated you to factor out \(x^{2}+x+1\) ?

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4) \( \deg(P^2) = 2 \deg(P) > 2 \deg Q = \deg(Q^2) \).

\(6 = \deg(x^6 + 1) = \deg(P^2 + Q^2) = \deg(P^2) = 2\deg(P) \).

\(\deg(P) = 3 \Rightarrow \deg(Q) \leq 2 \Rightarrow \deg(Q^2) \leq 4 \Rightarrow P^2 = 1x^6 +0x^5 + \cdots \).

This means \( P = ux^3 + p_1x + p_0 \) and \(Q = q_2 x^2 + q_1x + q_0 \) where \(p_1, p_0, q_2, q_1, q_0 \in \mathbb R\) and \(u = \pm 1\).

From \(P^2 + Q^2 = x^6 + 1 \) we get the following equations

\[ \displaystyle{ \cases{ p_0^2 + q_0^2 = 1 \\ \ \\ p_0p_1 + q_0q_1 = 0 \\ \ \\ p_1^2 + q_1^2 +2q_0q_2 = 0\\ \ \\ up_0 + q_1q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0 } } \]

The first one tell us there is \(\alpha \in [0, 2\pi) \) such that

\[ (p_0, q_0) = (\cos \alpha, \sin \alpha) \]

The second equation states that

\[ (p_0, q_0) \perp (p_1, q_1) \]

so must exist \( \rho \in \mathbb R \) such that

\[ (p_1, q_1) = (-\rho \sin \alpha, \rho \cos \alpha) \]

Assuming \( \cos\alpha \not= 0\) the fourth equation becomes

\[ q_2 = -\frac{u}{\rho} \]

Assuming \( \sin\alpha \not= 0\) the fifth and the third equations become

\[ q_2^2 = 2u\rho\sin\alpha \\ q_2 = - \frac{\rho^2}{2\sin\alpha} \]

Squaring the latter we get

\[ \frac{\rho^4}{4\sin^2\alpha} = q_2^2 = 2u\rho\sin\alpha\]

and so

\[ \rho^3 = 8u\sin^3\alpha \]

By the iniectivity of the function \(f(x) = x^3\) and by the obvious fact that \(u^3 = u\) we get

\[ \rho = 2u\sin\alpha\]

Using this equality into

\[ q_2 = -\dfrac{u}{\rho} \quad \textrm{ and } \quad q_2 = - \dfrac{\rho^2}{2\sin\alpha} \]

we see that

\[ q_2 = -2\sin\alpha \quad \textrm{ and } \quad \sin^2\alpha = \dfrac{1}{4} \]

So we find \(\sin\alpha = \pm \dfrac{1}{2}\) and the possible values for \( \alpha \) are

\[\alpha = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}\]

and the following identities

\[ \displaystyle{ \cases{ p_0 = \cos\alpha \\ \ \\ q_0 = \sin\alpha \\ \ \\ p_1 = -2u\sin^2\alpha = -2uq_o^2\\ \ \\ q_1 = 2u\sin\alpha\cos\alpha = 2uq_0p_0 \\ \ \\ q_2 = -2\sin\alpha = -2q_0 } } \]

Putting in the possible values for \(\alpha\) we get the following 8 couples of polynomials (each vale giving two polynomials depending on \(u\) being \(1\) or \(-1\) ) that we can write in short in the following way.

Define

\[ A =x^3 - \dfrac{x}{2} + \dfrac{\sqrt 3}{2} \quad \textrm{ and } \quad B = x^2 - \dfrac{\sqrt3}{2}x - \dfrac{1}{2} \] \[ C =x^3 - \dfrac{x}{2} - \dfrac{\sqrt 3}{2} \quad \textrm{ and } \quad D = x^2 + \dfrac{\sqrt3}{2}x - \dfrac{1}{2} \]

then the following are solution of our problem

\[ (P,Q) =\cases{ (\pm A , \pm B) \\ \ \\ (\pm C , \pm D) } \]

This solution left out the cases in which \(\cos \alpha \cdot \sin \alpha = 0 \). These cases lead to trivial solutions as we can easily find out.

If \(\cos\alpha = 0\) then \(p_0 = 0\) and \(q_0 = v = \pm1\), and we immediately get \(q_1 = 0\) and \(p_1 = -v\rho\). The equations \[\cases{p_1^2 + q_1^2 +2q_0q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0} \quad \quad \textrm{become} \quad \quad \cases{\rho^2 +2vq_2 = 0 \\ \ \\ -2uv\rho + q_2^2 = 0 } \]

\[\cases{\rho(\rho^3 -8uv) = 0 \\ \ \\ q_2= - \dfrac{\rho^2}{2v} } \]

By the upper equation we have either \(\rho = 0\) or \(\rho = 2uv\) If \(\rho = 0\) we get \(q_2=p_1=0\) so in this case

\[ \displaystyle{ \cases{ p_0 = 0 \\ \ \\ q_0 = v = \pm1 \\ \ \\ p_1 = 0\\ \ \\ q_1 = 0 \\ \ \\ q_2 = 0 } } \]

that leads to these four couple of solutions

\[ (P,Q) = (\pm x^3 , \pm 1) \]

If \(\rho = 2uv\) we get \(q_2= -2v, p_1= -2u \) so in this case

\[ \displaystyle{ \cases{ p_0 = 0 \\ \ \\ q_0 = v = \pm1 \\ \ \\ p_1 = -2u\\ \ \\ q_1 = 0 \\ \ \\ q_2 = -2v } } \]

that gives these other four couples

\[ (P,Q) = (\pm (x^3 -2x), \pm (2x^2 -1) ) \]

Last case is when \( \sin \alpha = 0 \). In this case we get \(q_0 = p_1 = 0, p_0 = v, q_1 = \rho v \) where \(v = \pm 1 \). The equations

\[\cases{p_1^2 + q_1^2 +2q_0q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0} \quad \quad \textrm{become} \quad \quad \cases{\rho^2 = 0 \\ \ \\ q_2^2 = 0 } \]

so \(\rho=q_2=q_1=0\) and the equation \(up_0 + q_1q_2 = 0 \) becomes

\[uv = 0 \]

that is always false. So this latter case doesn't lead to any further solution and our analysis is complete.

We found 16 pairs of polynomials \( (P, Q) \in (\mathbb R[x])^2 \) that satisfy the equation.

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It's not clear to me what you're trying to do here. Are you saying that no such polynomials exist?

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I don't know! I ended up finding 8 couples of polynomials. I was sure I posted the solution but I guess my internet connection (or me) failed to post the whole solution. I will repost it in short time. I'm writing right now.

-----EDIT---- Just wrote the solution that I made this morning. I just realized that I didn't examine some cases, and so my solution is not complete, but the cases left put are easy to handle (I feel) and I'll do as soon as possible.

--- EDIT --- Added also the trivial cases (boring). Now I think the problem is completely discussed and solved.

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A more straightforward solution for Q-2 using Calvin's hint:We first modify the expression of LHS by multiplying numerator and denominator of the expression inside brackets by \((1+\sin\theta+i\cos\theta)\).

\[\textrm{LHS}=\left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^n=\left(\frac{(1+\sin\theta+i\cos\theta)^2}{(1+\sin\theta)^2-(i\cos\theta)^2}\right)^n\\ \implies \textrm{LHS}=\left(\frac{1-\cos^2\theta+\sin^2\theta+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta}{1+\sin^2\theta+\cos^2\theta+2\sin\theta}\right)^n\\ \implies \textrm{LHS}=\left(\frac{2\sin\theta+2\sin^2\theta+2i\cos\theta+2i\cos\theta\sin\theta}{2+2\sin\theta}\right)^n\\ \implies \textrm{LHS}=\left(\frac{2(1+\sin\theta)(\sin\theta+i\cos\theta)}{2(1+\sin\theta)}\right)^n=(\sin\theta+i\cos\theta)^n\\ \implies \textrm{LHS}=\bigg(\cos\left(\frac{\pi}{2}-\theta\right)+i\sin\left(\frac{\pi}{2}-\theta\right)\bigg)^n\]

Using Euler's formula \(e^{i\theta}=\cos\theta+i\sin\theta\) and laws of indices, we can modify LHS as,

\[\textrm{LHS}=\large \left(e^{\left(\frac{\pi}{2}-\theta\right)}\right)^n=e^{n\left(\frac{\pi}{2}-\theta\right)}=e^{\left(\frac{n\pi}{2}-n\theta\right)}=\cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)=\textrm{RHS}\]

\[\therefore\quad \left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^{n} = \cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)\]

And we are done. \(_\square\)

Elementary identities used:Log in to reply

I also noticed that one can directly get the RHS form without going into Euler's formula by directly using De Moivre's Theorem.

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\(Q-3\)

This is not a solutionI Used Newtons sums to get answer \(\boxed{60}\)

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Did you mean Q3?

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Yes.

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There is a pretty simple solution.

Hint:What is \( (x-3)^4 \)?Log in to reply

Yes sir , on opening all the terms I got

\[P_4-12P_3+54P_2-108P_1+324\]

Where \(P_n=a^n+b^n+c^n+d^n\)

now we can apply newtons sums.

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That was quite a big hint. :D

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2) is pretty simple. First, replace \( \theta = \pi/2 - a \).

Then, \( 1 +\sin \theta + i \cos \theta = 1 + \cos a + i \sin a = 2\cos^2a/2 + i2\sin(a/2) \cos(a/2) = 2\cos(a/2)(\cos a/2+i\sin a/2) \)

Put \( b = a/2 \) Likewise, \( 1 + \sin\theta - i\cos \theta = 2\cos b(\cos b-i\sin b). \)

Then \( \dfrac{1 + \sin\theta + i \cos \theta }{1 + \sin\theta - i\cos \theta } = \dfrac{2\cos b(\cos b+i\sin b)}{2\cos b(\cos b-i\sin b)} = \dfrac{\cos b+i\sin b}{\cos b-i\sin b} \)

\( = \dfrac{(\cos b+i\sin b)}{(\cos b-i\sin b)} * \dfrac{(\cos b+i\sin b)}{(\cos b+i\sin b)} = \dfrac{(\cos b+i\sin b)^2}{\cos^2 b + \sin^2 b} = (\cos b+i\sin b)^2 \)

Therefore, \( (\dfrac{1 + \sin\theta + i \cos \theta }{1 + \sin\theta - i\cos \theta })^n = ((\cos b+i\sin b)^2)^n = (\cos b + i\sin b)^{2n} \)

\( = \cos 2bn + i\sin 2bn = \cos an + i\sin an = \cos(n(\pi/2 - a)) + i\sin(n(\pi/2 - a)) \)

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There's a more straightforward way to solve 2.

Hint:\( \left( e ^ { i \theta } \right) ^n = e^{i \theta n } \).Log in to reply

That was a big hint too. :P

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I'm confused? Isn't Demoivre's theroem equivalent to that fact?

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3) a,b,c,d are roots of equation , so

(a-3)^4 = 10a-15 ------m

(b-3)^4 = 10b-15 -------n

(c-3)^4 = 10c-15 -------o

(d-3)^4 = 10d-15 -------p

from vieta's formula we'll get a+b+c+d = 12

m+n+o+p ; (a-3)^4+(b-3)^4+(c-3)^4+(d-3)^4 = 10(a+b+c+d)-60 = 10(12)-60 = 60

so, (a-3)^4+(b-3)^4+(c-3)^4+(d-3)^4 = 60

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\(1.\quad { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }-{ x }^{ 2 }-x-1\\ ={ x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }-\left( { x }^{ 2 }+x+1 \right) \\ ={ x }^{ 3 }\left( { x }^{ 2 }+x+1 \right) -1\left( { x }^{ 2 }+x+1 \right) \\ =\left( { x }^{ 2 }+x+1 \right) \left( { x }^{ 3 }-1 \right) \\ =\left( { x }^{ 2 }+x+1 \right) \left( { x }^{ 2 }+x+1 \right) \left( x-1 \right) \\ \\ Solving\quad quadratics,\quad we\quad get\quad roots\quad 1,\omega ,\omega ,{ \omega }^{ 2 },{ \omega }^{ 2 }\)

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