Write a full solution,
Find all roots of \(x^{5}+x^{4}+x^{3}-x^{2}-x-1 = 0\).
Prove that (1+sinθ−icosθ1+sinθ+icosθ)n=cos(2nπ−nθ)+isin(2nπ−nθ)
If a,b,c,d are roots of equation x4−12x3+54x2−118x+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=x6+1.
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Thailand Math POSN 2nd round 2015
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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 ∀ x∈{a,b,c,d}⟹(x−3)4=10x−15 ∀ x∈{a,b,c,d}
Using the last result, we have the sum as,
x∈{a,b,c,d}∑(x−3)4=x∈{a,b,c,d}∑(10x−15)
Using Vieta's formulas, we have,
x∈{a,b,c,d}∑(x)=12⟹x∈{a,b,c,d}∑(10x)=120
Hence, the required sum evaluates as,
x∈{a,b,c,d}∑(x−3)4=x∈{a,b,c,d}∑(10x−15)=⎩⎨⎧⎝⎛x∈{a,b,c,d}∑(10x)⎠⎞−⎝⎛x∈{a,b,c,d}∑(15)⎠⎞⎭⎬⎫=120−15×4=120−60=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. :)
Q−1
x5+x4+x3−x2−x−1=(x−1)(x4+2x3+3x2+2x+10)⟹one root is 1
x4+2x3+3x2+2x+1=0⟹x2+2x+3+x2+x21=0 ⟹(x+x1)2−2+2(x+x1)+3=0
x+x1=m
m2+2m+1=0⟹(m+1)2=0⟹m=1,1
x+x1=1On solving using quadratic equations we get x=ω,ω2
(⟹we get 4 roots from here ω,ω,ω2,ω2)
roots are ⟹ ω,ω2,1,ω,ω2
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You can factor it the easier way: x5+x4+x3−x2−x−1=x3(x2+x2+1)−(x2+x+1)=(x3−1)(x2+x+1)
⇒(x−1)(x2+x+1)2=0⇒x=1,ω,ω2
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What motivated you to factor out x2+x+1 ?
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x2+x+1 is a factor, so I just factor x2+x+1 from the start.
Looking at Parth's solution tells us that4) deg(P2)=2deg(P)>2degQ=deg(Q2).
6=deg(x6+1)=deg(P2+Q2)=deg(P2)=2deg(P).
deg(P)=3⇒deg(Q)≤2⇒deg(Q2)≤4⇒P2=1x6+0x5+⋯.
This means P=ux3+p1x+p0 and Q=q2x2+q1x+q0 where p1,p0,q2,q1,q0∈R and u=±1.
From P2+Q2=x6+1 we get the following equations
⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧p02+q02=1 p0p1+q0q1=0 p12+q12+2q0q2=0 up0+q1q2=0 2up1+q22=0
The first one tell us there is α∈[0,2π) such that
(p0,q0)=(cosα,sinα)
The second equation states that
(p0,q0)⊥(p1,q1)
so must exist ρ∈R such that
(p1,q1)=(−ρsinα,ρcosα)
Assuming cosα=0 the fourth equation becomes
q2=−ρu
Assuming sinα=0 the fifth and the third equations become
q22=2uρsinαq2=−2sinαρ2
Squaring the latter we get
4sin2αρ4=q22=2uρsinα
and so
ρ3=8usin3α
By the iniectivity of the function f(x)=x3 and by the obvious fact that u3=u we get
ρ=2usinα
Using this equality into
q2=−ρu and q2=−2sinαρ2
we see that
q2=−2sinα and sin2α=41
So we find sinα=±21 and the possible values for α are
α=6π,65π,67π,611π
and the following identities
⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧p0=cosα q0=sinα p1=−2usin2α=−2uqo2 q1=2usinαcosα=2uq0p0 q2=−2sinα=−2q0
Putting in the possible values for α 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=x3−2x+23 and B=x2−23x−21 C=x3−2x−23 and D=x2+23x−21
then the following are solution of our problem
(P,Q)=⎩⎪⎨⎪⎧(±A,±B) (±C,±D)
This solution left out the cases in which cosα⋅sinα=0. These cases lead to trivial solutions as we can easily find out.
If cosα=0 then p0=0 and q0=v=±1, and we immediately get q1=0 and p1=−vρ. The equations ⎩⎪⎨⎪⎧p12+q12+2q0q2=0 2up1+q22=0become⎩⎪⎨⎪⎧ρ2+2vq2=0 −2uvρ+q22=0
⎩⎪⎪⎨⎪⎪⎧ρ(ρ3−8uv)=0 q2=−2vρ2
By the upper equation we have either ρ=0 or ρ=2uv If ρ=0 we get q2=p1=0 so in this case
⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧p0=0 q0=v=±1 p1=0 q1=0 q2=0
that leads to these four couple of solutions
(P,Q)=(±x3,±1)
If ρ=2uv we get q2=−2v,p1=−2u so in this case
⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧p0=0 q0=v=±1 p1=−2u q1=0 q2=−2v
that gives these other four couples
(P,Q)=(±(x3−2x),±(2x2−1))
Last case is when sinα=0. In this case we get q0=p1=0,p0=v,q1=ρv where v=±1. The equations
⎩⎪⎨⎪⎧p12+q12+2q0q2=0 2up1+q22=0become⎩⎪⎨⎪⎧ρ2=0 q22=0
so ρ=q2=q1=0 and the equation up0+q1q2=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)∈(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.
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+sinθ+icosθ).
LHS=(1+sinθ−icosθ1+sinθ+icosθ)n=((1+sinθ)2−(icosθ)2(1+sinθ+icosθ)2)n⟹LHS=(1+sin2θ+cos2θ+2sinθ1−cos2θ+sin2θ+2sinθ+2icosθ+2icosθsinθ)n⟹LHS=(2+2sinθ2sinθ+2sin2θ+2icosθ+2icosθsinθ)n⟹LHS=(2(1+sinθ)2(1+sinθ)(sinθ+icosθ))n=(sinθ+icosθ)n⟹LHS=(cos(2π−θ)+isin(2π−θ))n
Using Euler's formula eiθ=cosθ+isinθ and laws of indices, we can modify LHS as,
LHS=(e(2π−θ))n=en(2π−θ)=e(2nπ−nθ)=cos(2nπ−nθ)+isin(2nπ−nθ)=RHS
∴(1+sinθ−icosθ1+sinθ+icosθ)n=cos(2nπ−nθ)+isin(2nπ−nθ)
And we are done. □
Elementary identities used:
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I also noticed that one can directly get the RHS form without going into Euler's formula by directly using De Moivre's Theorem.
Q−3
This is not a solution
I Used Newtons sums to get answer 60
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Did you mean Q3?
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Yes.
There is a pretty simple solution.
Hint: What is (x−3)4?
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Yes sir , on opening all the terms I got
P4−12P3+54P2−108P1+324
Where Pn=an+bn+cn+dn
now we can apply newtons sums.
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That was quite a big hint. :D
2) is pretty simple. First, replace θ=π/2−a.
Then, 1+sinθ+icosθ=1+cosa+isina=2cos2a/2+i2sin(a/2)cos(a/2)=2cos(a/2)(cosa/2+isina/2)
Put b=a/2 Likewise, 1+sinθ−icosθ=2cosb(cosb−isinb).
Then 1+sinθ−icosθ1+sinθ+icosθ=2cosb(cosb−isinb)2cosb(cosb+isinb)=cosb−isinbcosb+isinb
=(cosb−isinb)(cosb+isinb)∗(cosb+isinb)(cosb+isinb)=cos2b+sin2b(cosb+isinb)2=(cosb+isinb)2
Therefore, (1+sinθ−icosθ1+sinθ+icosθ)n=((cosb+isinb)2)n=(cosb+isinb)2n
=cos2bn+isin2bn=cosan+isinan=cos(n(π/2−a))+isin(n(π/2−a))
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There's a more straightforward way to solve 2.
Hint: (eiθ)n=eiθn.
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That was a big hint too. :P
I'm confused? Isn't Demoivre's theroem equivalent to that fact?
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(cos(2π−θ)+sin(2π−θ))n, which would explain why we did the calculations that we did.
Ah, the difference is more in terms of motivation. IE, to explain that the intermediate step is3) 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
1.x5+x4+x3−x2−x−1=x5+x4+x3−(x2+x+1)=x3(x2+x+1)−1(x2+x+1)=(x2+x+1)(x3−1)=(x2+x+1)(x2+x+1)(x−1)Solvingquadratics,wegetroots1,ω,ω,ω2,ω2