# Algebra (Thailand Math POSN 2nd round)

Write a full solution,

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

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

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

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

Note by Samuraiwarm Tsunayoshi
5 years, 11 months ago

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Solution 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,

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

- 5 years, 11 months ago

Nicely done.

Staff - 5 years, 11 months ago

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

- 5 years, 11 months ago

$Q-1$

$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 $\implies$ $\boxed{\omega,\omega^2,1,\omega,\omega^2}$

- 5 years, 11 months ago

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

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

- 5 years, 11 months ago

What motivated you to factor out $x^{2}+x+1$ ?

- 4 years, 9 months ago

Looking at Parth's solution tells us that $x^2 + x+ 1$ is a factor, so I just factor $x^2+x+1$ from the start.

- 4 years, 9 months ago

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.

- 5 years, 11 months ago

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

Staff - 5 years, 11 months ago

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.

- 5 years, 11 months ago

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:

• $\cos^2\theta+\sin^2\theta=1$
• $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
• $\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta$
• $\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta$

- 5 years, 11 months ago

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

- 5 years, 11 months ago

$Q-3$

This is not a solution

I Used Newtons sums to get answer $\boxed{60}$

- 5 years, 11 months ago

Did you mean Q3?

- 5 years, 11 months ago

Yes.

- 5 years, 11 months ago

There is a pretty simple solution.

Hint: What is $(x-3)^4$?

Staff - 5 years, 11 months ago

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.

- 5 years, 11 months ago

See Prasun's solution at the top for the "two-line" solution to this problem.

Staff - 5 years, 11 months ago

That was quite a big hint. :D

- 5 years, 11 months ago

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

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

- 5 years, 11 months ago

There's a more straightforward way to solve 2.

Hint: $\left( e ^ { i \theta } \right) ^n = e^{i \theta n }$.

Staff - 5 years, 11 months ago

That was a big hint too. :P

- 5 years, 11 months ago

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

- 5 years, 11 months ago

Yes it is. The hints that I give are often to suggest alternative approaches, that you would have to work out how to apply them. In this case, see Prasun's solution above.

Staff - 5 years, 11 months ago

I was asking because I had used Demoivre's theorem. I'm also not sure how Prasun's solution is vastly different from mine, since we use almost the same approach. The only major difference in our solutions is the way we simplify the original expression

- 5 years, 11 months ago

Ah, the difference is more in terms of motivation. IE, to explain that the intermediate step is $\left( \cos ( \frac{\pi}{2} - \theta) + \sin ( \frac{\pi}{2} - \theta ) \right) ^ n$, which would explain why we did the calculations that we did.

Staff - 5 years, 11 months ago

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

- 5 years, 10 months ago

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

- 5 years, 11 months ago