1, 2, 3, _ ?

Algebra Level 4

x , y x, y and z z are complex numbers satisfying

{ x 1 + y 1 + z 1 = 1 x 2 + y 2 + z 2 = 2 x 3 + y 3 + z 3 = 3 \begin{cases} x^1+y^1+z^1 & = 1\\ x^2 + y^2 + z^2 & = 2 \\ x^3 + y^3 + z^3 & = 3 \\ \end{cases}

The value of x 4 + y 4 + z 4 x^4 + y^4 + z^4 can be expressed as a b \frac {a}{b} , where a a and b b are positive coprime integers. What is the value of a + b a +b ?

This problem is proposed by Harshit .


The answer is 31.

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11 solutions

José Neto
May 20, 2014

As a way of simplicity we can call:
x 3 + y 3 + z 3 = S 3 = 3 x^3 + y^3 + z^3 = S_3 = 3 ,
x 2 + y 2 + z 2 = S 2 = 2 x^2 + y^2 + z^2 = S_2 = 2 ,
x 1 + y 1 + z 1 = S 1 = 1 x^1 + y^1 + z^1 = S_1 = 1 ,
x 4 + y 4 + z 4 = S 4 x^4 + y^4 + z^4 = S_4 .


We can create our own polynomial wich the roots are x, y, z. So we have that:

x 3 + a . x 2 + b . x + c = 0 , y 3 + a . y 2 + b . y + c = 0 , z 3 + a . z 2 + b . z + c = 0 x^3 + a.x^2 + b.x + c = 0, y^3 + a.y^2 + b.y + c = 0, z^3 + a.z^2 + b.z + c = 0

Then we sum these equations above, we find that:

( x 3 + y 3 + z 3 ) + a . ( x 2 + y 2 + z 2 ) + b . ( x + y + z ) + 3 c = 0 (x^3 + y^3 + z^3) + a.(x^2 + y^2 + z^2) + b.(x + y + z) + 3c = 0

So: S 3 + a . S 2 + b . S 1 + 3 c = 0 S_3 + a.S_2 + b.S_1 + 3c = 0 , 3 + 2 a + b + 3 c = 0 3 + 2a + b + 3c = 0 .

And from the polynomials properties (Vieta) we get that:

a = ( x + y + z ) = 1 a = - (x + y + z) = -1 , b = x y + y z + z x = ( 1 / 2 ) [ ( x + y + z ) 2 ( x 2 + y 2 + z 2 ) ] = 1 / 2 b = xy + yz + zx = (1/2)[(x + y + z)^2 - (x^2 + y^2 + z^2)] = -1/2

So: 3 + 2 a + b + 3 c = 0 , 3 2 1 / 2 + 3 c = 0 , c = 1 / 6 3 + 2a + b + 3c = 0, 3 - 2 - 1/2 + 3c = 0, c = -1/6 .

Now we can do this:

x 3 + a . x 2 + b . x + c = 0 x^3 + a.x^2 + b.x + c = 0

Multiplying by x:

x 4 + a . x 3 + b . x 2 + c . x = 0 x^4 + a.x^3 + b.x^2 + c.x = 0

Doing the same to y and z, we sum them all and get to:

S 4 + a . S 3 + b . S 2 + c . S 1 = 0 S_4 + a.S_3 + b.S_2 + c.S_1 = 0
S 4 3 1 1 / 6 = 0 S_4 - 3 - 1 - 1/6 = 0
S 4 = 25 / 6 S_4 = 25/6

Hence, we have 25 + 6 = 31 25+6=31 .

There are many algebraic identities that can be used to arrive at the result. Newton's Identities provides a quick way to calculate S n S_n . Note that solving the recurrence is equivalent to finding the values of x , y , z x, y, z .

Calvin Lin Staff - 7 years ago

You create three polynomials of which the roots are x y and z. How do we know that the coefficients a b and c in these polynomials are the same? (I dare say it's obvious to a proper mathematician but I'm not!)

Richard Chester-Browne - 5 years, 1 month ago
Wei Liang Gan
May 20, 2014

x y + y z + z x = 1 2 ( ( x + y + z ) 2 ( x 2 + y 2 + z 2 ) ) = 1 2 ( 1 2 2 ) = 1 2 xy+yz+zx=\frac{1}{2}((x+y+z)^2-(x^2+y^2+z^2))=\frac{1}{2}(1^2-2)=-\frac{1}{2} x 4 + y 4 + z 4 = 1 3 ( 6 ( x y + y z + z x ) 2 + 4 ( x 3 + y 3 + z 3 ) ( x + y + z ) ( x + y + z ) 4 ) = 1 3 ( 6 ( 1 2 ) 2 + 4 ( 3 ) ( 1 ) 1 4 ) = 25 6 x^4+y^4+z^4=\frac{1}{3}(6(xy+yz+zx)^2+4(x^3+y^3+z^3)(x+y+z)-(x+y+z)^4)=\frac{1}{3}(6(-\frac{1}{2})^2+4(3)(1)-1^4)=\frac{25}{6} Therefore, the required answer is 25 + 6 = 31 25+6=31

Jo Ong
May 20, 2014

(x + y + z)^2 = (x^2 + y^2 + z^2) + 2(xy + yz + zx) (xy + yz + zx) = -1/2 Now, x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = 5/2 3xyz = 1/2 xyz = 1/6 (x^2 + y^2 + z^2)^2 = (x^4 + y^4 + z^4) + 2(x^2y^2 + y^2z^2 + z^2x^2) ....(1) x^2y^2 + y^2z^2 + z^2x^2 = (xy + yz + zx)^2 - 2xyz(x + y + z) = -1/12 From (1), x^4 + y^4 + z^4 = 25/6

Natsir Muhammad
May 20, 2014

x^2+y^2+z^2 = (x+y+z)^2 - 2(xy+xz+yz) 2 = 1 - 2(xy+xz+yz)

xy+xz+yz = -1/2

x^3+y^3+z^3 = (x+y+z)^3 - [ 3 (xy+xz+yz)(x+y+z) - 3xyz ] 3 = 1 - [ 3 . -1/2 . 1 - 3xyz ] 3 = 1 + 3/2 + 3xyz

xyz = 1/6

x^4 + y^4 + z^4 = (x^2+y^2+z^2)^2 - [ 2 [ (xy)^2 + (xz)^2 + (yz)^2 ] ] = 4 - [ 2 [ (xy+xz+yz)^2 - 2 xyz(x+y+z) ] ] = 4 - [ 2 [ 1/4 - 1/3 ] ] = 4 + 1/6 = 25 / 6

so, a+b = 31

Brian Reinhart
May 20, 2014

Squaring the first equation and substituting the second gives us ( x + y + z ) 2 = x 2 + y 2 + z 2 + 2 ( x y + x z + y z ) = 2 + 2 ( x y + x z + y z ) = 1 (x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=2+2(xy+xz+yz)=1 , so 1 + x y + x z + y z = 1 2 1+xy+xz+yz=\frac{1}{2} and x y + x z + y z = 1 2 xy+xz+yz=-\frac{1}{2} . Multiplying x + y + z x+y+z by x y + x z + y z xy+xz+yz gives us x 2 y + x 2 z + y 2 x + y 2 z + z 2 x + z 2 y + 3 x y z = 1 1 2 = 1 2 x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+3xyz=1*-\frac{1}{2}=-\frac{1}{2} . Rearranging x + y + z = 1 x+y+z=1 and using the other two equations gives us x 2 ( y + z ) + y 2 ( x + z ) + z 2 ( y + z ) + 3 x y z = x 2 ( 1 x ) + y 2 ( 1 y ) + z 2 ( 1 z ) + 3 x y z = x 2 + y 2 + z 2 ( x 3 + y 3 + z 3 ) + 3 x y z = 2 3 + 3 x y z = 3 x y z 1 = 1 2 x^2(y+z)+y^2(x+z)+z^2(y+z)+3xyz=x^2(1-x)+y^2(1-y)+z^2(1-z)+3xyz=x^2+y^2+z^2-(x^3+y^3+z^3)+3xyz=2-3+3xyz=3xyz-1=-\frac{1}{2} . Isolating x y z xyz tells us x y z = 1 6 xyz=\frac{1}{6} . Just a little more: Squaring x y + x z + y z xy+xz+yz gives us ( x y ) 2 + ( y z ) 2 + ( x z ) 2 + 2 ( x y z ) ( x + y + z ) = ( x y ) 2 + ( y z ) 2 + ( x z ) 2 + 2 1 6 1 = ( x y ) 2 + ( y z ) 2 + ( x z ) 2 + 1 3 = 1 4 (xy)^2+(yz)^2+(xz)^2+2(xyz)(x+y+z)=(xy)^2+(yz)^2+(xz)^2+2*\frac{1}{6}*1=(xy)^2+(yz)^2+(xz)^2+\frac{1}{3}=\frac{1}{4} . Thus, ( x y ) 2 + ( x z ) 2 + ( y z ) 2 = 1 12 (xy)^2+(xz)^2+(yz)^2=-\frac{1}{12} . We're almost there: Squaring x 2 + y 2 + z 2 x^2+y^2+z^2 gives us x 4 + y 4 + z 4 + 2 ( ( x y ) 2 + ( x z ) 2 + ( y z ) 2 ) = x 4 + y 4 + z 4 1 6 = 4 x^4+y^4+z^4+2((xy)^2+(xz)^2+(yz)^2)=x^4+y^4+z^4-\frac{1}{6}=4 , and we have that x 4 + y 4 + z 4 = 4 + 1 6 = 25 6 x^4+y^4+z^4=4+\frac{1}{6}=\frac{25}{6} , for a final answer of 25 + 6 = 31 25+6=31

Note: One could easily use the Newton-Girard formulas to solve this problem. However, for those too lazy to remember them/have never heard of them, this is a solution which actually derives the cubic equation (use Vieta's with the x+y+z, xy+xz+yz, xyz) and then goes on to use these to find the fourth power. I believe that in a similar fashion you can derive some of the N-G formulae

(the LaTeX is not rendering in my solution, at least not when I preview it. I have tried to put the necessary LaTeX parentheses, but please feel free to LaTeX edit this as much as you please. Perhaps it is something to do with my mac, or something to do with my syntax, but it renders in the problem and not the solution. Could you possibly inform me of a reason for the failure to render?)

Zi Song Yeoh
May 20, 2014

Let a , b , c a, b, c be the roots of the polynomial x 3 + d x 2 + e x + f = 0 x^{3} + dx^{2} + ex + f = 0 . Let S n = a n + b n + c n S_{n} = a^{n} + b^{n} + c^{n} . By Vieta's Formula, d = 1 d = -1 . And 2 e = 2 ( a b + b c + a c ) = S 1 2 S 2 = 1 e = 1 2 2e = 2(ab + bc + ac) =S_{1}^{2} - S_{2} = -1 \Rightarrow e = \frac{1}{2} . Note that a n + 3 a n + 2 + 1 2 a n + 1 + f a n a^{n + 3} - a^{n + 2} + \frac{1}{2}a_{n + 1} + fa_{n} . Adding up similar equations yields S n + 3 S n + 2 + 1 2 S n + 1 + f S n = 0 S_{n + 3} - S_{n + 2} + \frac{1}{2}S_{n + 1} + fS_{n} = 0 . Substituting n = 0 n = 0 gives 3 2 + 1 2 + 3 f = 0 f = 1 6 3 - 2 + \frac{1}{2} + 3f = 0 \Rightarrow f = -\frac{1}{6} . Then substituting n = 1 n = 1 gives S 4 = 25 6 S_{4} = \frac{25}{6} . So, a + b = 31 a + b = 31 .

Akshay Ravikumar
May 20, 2014

Because ( x + y + z ) 2 = 1 = x 2 + y 2 + z 2 + 2 ( x y + y z + x z ) = 2 + 2 ( x y + y z + x z ) (x+y+z)^2=1=x^2+y^2+z^2+2(xy+yz+xz)=2+2(xy+yz+xz) , we have xy+yz+xz=- \dfrac[1}{2}

In addition, because (

Lawrence Limesa
May 20, 2014

1st, square the first equation ( x + y + z ) 2 = x 2 + y 2 + z 2 + 2 x y + 2 y z + 2 x z (x+y+z)^{2} = x^2 + y^2 + z^2 + 2xy + 2yz + 2xz getting the result of 2 x y + 2 y z + 2 z x = 1 2xy + 2yz + 2zx = -1

2nd, multiply the first and the second equation and you will get x y z = 1 6 xyz = \frac{1}{6}

Then multiply the first and the third equation ( x + y + z ) ( x 3 + y 3 + z 3 ) = x 4 + y 4 + z 4 + x y ( x 2 + y 2 ) + x z ( x 2 + z 2 ) + y z ( y 2 + z 2 ) (x+y+z)(x^3 + y^3 + z^3) = x^4 + y^4 + z^4 + xy ( x^2 + y^2) + xz ( x^2 + z^2) + yz ( y^2 + z^2 ) by subtituing x 2 + y 2 x^2 + y^2 with 2 z 2 2- z^2 and subtituting the other sum of squares we can get that 25 6 = x 4 + y 4 + z 4 \frac{25}{6} = x^4 + y^4 + z^4

Kelvin Hong
Jul 7, 2018

Seems that this question is from a long time ago, but I think I should give this simple notation, just reduce some time where we are writing a bunch of x , y , z x,y,z .

Let S 1 = x + y + z , S 2 = x y + y z + z x , S 3 = x y z S_1=x+y+z, S_2=xy+yz+zx, S_3=xyz , also P n = x n + y n + z n P_n=x^n+y^n+z^n , we have S 1 = P 1 = 1 S 2 = ? , P 2 = 2 S 3 = ? , P 3 = 3 \begin{aligned}S_1=P_1&=1\\S_2=? \quad&,P_2=2\\S_3=? \quad&,P_3=3\end{aligned} then we need to find P 4 P_4 .

By using Newton's Identity several times: P n = S 1 P n 1 S 2 P n 2 + S 3 P n 3 P_n=S_1P_{n-1}-S_2P_{n-2}+S_3P_{n-3} , we deduce that P 2 = S 1 2 2 S 2 2 = 1 2 S 2 S 2 = 1 2 \begin{aligned}P_2&=S_1^2-2S_2\\2&=1-2S_2\\\therefore S_2&=-\frac12\end{aligned} Compute again: P 3 = S 1 P 2 S 2 P 1 + 3 S 3 3 = 2 + 1 2 + 3 S 3 S 3 = 1 6 \begin{aligned}P_3&=S_1P_2-S_2P_1+3S_3\\3&=2+\frac12+3S_3\\\therefore S_3&=\frac16\end{aligned} Now we can find out P 4 = S 1 P 3 S 2 P 2 + S 3 P 1 = 3 + 1 + 1 6 = 25 6 \begin{aligned}P_4&=S_1P_3-S_2P_2+S_3P_1\\&=3+1+\frac16\\&=\frac{25}6\end{aligned} the answer is 25 + 6 = 31 25+6=\boxed{31} .

Small note: Maybe some of you don't agree with me, but the identity P 2 = S 1 P 1 S 2 P 0 + S 3 P 1 P_2=S_1P_1-S_2P_0+S_3P_{-1} is correct! If calculate from RHS, we have S 1 P 1 S 2 P 0 + S 3 P 1 = S 1 2 S 2 ( x 0 + y 0 + z 0 ) + x y z ( 1 / x + 1 / y + 1 / z ) = ( x + y + z ) 2 3 S 2 + x y + y z + z x = ( x + y + z ) 2 2 S 2 = x 2 + y 2 + z 2 = P 2 \begin{aligned}S_1P_1-S_2P_0+S_3P_{-1}&=S_1^2-S_2(x^0+y^0+z^0)+xyz(1/x+1/y+1/z)\\&=(x+y+z)^2-3S_2+xy+yz+zx\\&=(x+y+z)^2-2S_2\\&=x^2+y^2+z^2\\&=P_2\end{aligned} It means that P 0 P_0 and P 1 P_{-1} is making sense.

Hadia Qadir
Sep 7, 2015

x^2+y^2+z^2 = (x+y+z)^2 - 2(xy+xz+yz) 2 = 1 - 2(xy+xz+yz) xy+xz+yz = -1/2 x^3+y^3+z^3 = (x+y+z)^3 - [ 3 (xy+xz+yz)(x+y+z) - 3xyz ] 3 = 1 - [ 3 . -1/2 . 1 - 3xyz ] 3 = 1 + 3/2 + 3xyz xyz = 1/6 x^4 + y^4 + z^4 = (x^2+y^2+z^2)^2 - [ 2 [ (xy)^2 + (xz)^2 + (yz)^2 ] ] = 4 - [ 2 [ (xy+xz+yz)^2 - 2 xyz(x+y+z) ] ] = 4 - [ 2 [ 1/4 - 1/3 ] ] = 4 + 1/6 = 25 / 6 so, a+b = 31

A simple way is to use Newton's Sum of powers !!

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