x , y and z are complex numbers satisfying
⎩ ⎪ ⎨ ⎪ ⎧ x 1 + y 1 + z 1 x 2 + y 2 + z 2 x 3 + y 3 + z 3 = 1 = 2 = 3
The value of x 4 + y 4 + z 4 can be expressed as b a , where a and b are positive coprime integers. What is the value of a + b ?
This problem is proposed by Harshit .
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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 . Note that solving the recurrence is equivalent to finding the values of x , y , z .
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!)
x y + y z + z x = 2 1 ( ( x + y + z ) 2 − ( x 2 + y 2 + z 2 ) ) = 2 1 ( 1 2 − 2 ) = − 2 1 x 4 + y 4 + z 4 = 3 1 ( 6 ( x y + y z + z x ) 2 + 4 ( x 3 + y 3 + z 3 ) ( x + y + z ) − ( x + y + z ) 4 ) = 3 1 ( 6 ( − 2 1 ) 2 + 4 ( 3 ) ( 1 ) − 1 4 ) = 6 2 5 Therefore, the required answer is 2 5 + 6 = 3 1
(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
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
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 , so 1 + x y + x z + y z = 2 1 and x y + x z + y z = − 2 1 . Multiplying x + y + z by x y + x z + y z 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 ∗ − 2 1 = − 2 1 . Rearranging 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 = − 2 1 . Isolating x y z tells us x y z = 6 1 . Just a little more: Squaring x y + x z + y z 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 ∗ 6 1 ∗ 1 = ( x y ) 2 + ( y z ) 2 + ( x z ) 2 + 3 1 = 4 1 . Thus, ( x y ) 2 + ( x z ) 2 + ( y z ) 2 = − 1 2 1 . We're almost there: Squaring 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 − 6 1 = 4 , and we have that x 4 + y 4 + z 4 = 4 + 6 1 = 6 2 5 , for a final answer of 2 5 + 6 = 3 1
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?)
Let a , b , c be the roots of the polynomial x 3 + d x 2 + e x + f = 0 . Let S n = a n + b n + c n . By Vieta's Formula, d = − 1 . And 2 e = 2 ( a b + b c + a c ) = S 1 2 − S 2 = − 1 ⇒ e = 2 1 . Note that a n + 3 − a n + 2 + 2 1 a n + 1 + f a n . Adding up similar equations yields S n + 3 − S n + 2 + 2 1 S n + 1 + f S n = 0 . Substituting n = 0 gives 3 − 2 + 2 1 + 3 f = 0 ⇒ f = − 6 1 . Then substituting n = 1 gives S 4 = 6 2 5 . So, a + b = 3 1 .
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 ) , we have xy+yz+xz=- \dfrac[1}{2}
In addition, because (
1st, square the first equation ( x + y + z ) 2 = x 2 + y 2 + z 2 + 2 x y + 2 y z + 2 x z getting the result of 2 x y + 2 y z + 2 z x = − 1
2nd, multiply the first and the second equation and you will get x y z = 6 1
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 ) by subtituing x 2 + y 2 with 2 − z 2 and subtituting the other sum of squares we can get that 6 2 5 = x 4 + y 4 + z 4
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 .
Let S 1 = x + y + z , S 2 = x y + y z + z x , S 3 = x y z , also P n = x n + y n + z n , we have S 1 = P 1 S 2 = ? S 3 = ? = 1 , P 2 = 2 , P 3 = 3 then we need to find 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 , we deduce that P 2 2 ∴ S 2 = S 1 2 − 2 S 2 = 1 − 2 S 2 = − 2 1 Compute again: P 3 3 ∴ S 3 = S 1 P 2 − S 2 P 1 + 3 S 3 = 2 + 2 1 + 3 S 3 = 6 1 Now we can find out P 4 = S 1 P 3 − S 2 P 2 + S 3 P 1 = 3 + 1 + 6 1 = 6 2 5 the answer is 2 5 + 6 = 3 1 .
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 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 It means that P 0 and P − 1 is making sense.
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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As a way of simplicity we can call:
x 3 + y 3 + z 3 = S 3 = 3 ,
x 2 + y 2 + z 2 = S 2 = 2 ,
x 1 + y 1 + z 1 = S 1 = 1 ,
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
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
So: S 3 + a . S 2 + b . S 1 + 3 c = 0 , 3 + 2 a + b + 3 c = 0 .
And from the polynomials properties (Vieta) we get that:
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
So: 3 + 2 a + b + 3 c = 0 , 3 − 2 − 1 / 2 + 3 c = 0 , c = − 1 / 6 .
Now we can do this:
x 3 + a . x 2 + b . x + c = 0
Multiplying by x:
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 − 3 − 1 − 1 / 6 = 0
S 4 = 2 5 / 6
Hence, we have 2 5 + 6 = 3 1 .