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Subject: "Sum of fourth powers"     Previous Topic | Next Topic
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NIH
Member since Jan-21-04
Mar-08-04, 06:19 PM (EST)
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"Sum of fourth powers"
 
   Here's an intriguing puzzle. It's quite challenging, but uses nothing beyond high school math.

The sum of three numbers is 6, the sum of their squares is 8, and the sum of their cubes is 5. What is the sum of their fourth powers?

Nick


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  Subject     Author     Message Date     ID  
Sum of fourth powers NIH Mar-08-04 TOP
  RE: Sum of fourth powers alexb Mar-08-04 1
  RE: Sum of fourth powers Naftali Mar-09-04 2
     RE: Sum of fourth powers NIH Mar-10-04 3
         RE: Sum of fourth powers Graham C Mar-12-04 4
             RE: Sum of fourth powers NIH Mar-15-04 5
                 RE: Sum of fourth powers Graham C Mar-17-04 6
                     RE: Sum of fourth powers NIH Mar-17-04 7
                         RE: Sum of fourth powers Graham C Mar-18-04 8

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alexb
Charter Member
1230 posts
Mar-08-04, 09:39 PM (EST)
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1. "RE: Sum of fourth powers"
In response to message #0
 
   It becomes much easier after a substitution

a = 2 + x
b = 2 + y
c = 2 + z

Then

x + y + z = 0
x2 + y2 + z2 = -4
x3 + y3 + z3 = 5,

if I am not wrong. An interesting expression for the sum of squares. Ignoring this, consider successively

(x + y + z)2, (x + y + z)3, and (x + y + z)4.


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Naftali
guest
Mar-09-04, 07:52 PM (EST)
 
2. "RE: Sum of fourth powers"
In response to message #0
 
   The solution is based on forming homogenous expressions(a+b+c)^4, (a+b+c)^2*(a^2+b^2+c^2), (a+b+c)*(a^3+b^3+c^3) and (a^2+b^2+c^2)^2, and then taking a weighted sum of those in which all terms but a^4+b^4+c^4 cancel out. It turns out to be the expression:


a^4+b^4+c^4 =
((a+b+c)^4 +8*(a+b+c)*(a^3+b^3+c^3) +3*(a^2+b^2+c^2)^2 -6*(a+b+c)^2*(a^2+b^2+c^2))/6

Now, substituting the numbers in the problem into the solution yields

(6^4+8*6*5+3*8^2-6*6^2*8)/6 = 0

which seems as a mistake in the values in the problem in that the only solution in real numbers to a^4+b^4+c^4=0 is a=0, b=0, c=0, which does not match the other data in the problem.

Best Regards,

Naftali

naftali.chayat@alvarion.com


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NIH
Member since Jan-21-04
Mar-10-04, 05:10 PM (EST)
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3. "RE: Sum of fourth powers"
In response to message #2
 
   Who said the numbers were real...?!

Nick


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Graham C
Member since Feb-5-03
Mar-12-04, 11:03 PM (EST)
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4. "RE: Sum of fourth powers"
In response to message #3
 
   >Who said the numbers were real...?!
>
>Nick

Following that somewhat cryptic comment, consider the simpler similar case of the two equations a+b=6, a^2+b^2=10, and find the value of a^3+b^3.

The two equations have no real solutions, but a^3+b^3 nevertheless equals -18.

(a = 3 + 2i & b = 3 - 2i -- or the other way round.)


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NIH
Member since Jan-21-04
Mar-15-04, 07:05 PM (EST)
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5. "RE: Sum of fourth powers"
In response to message #4
 
   Taking Graham's example, above, considering sum and product of a quadratic equation, a and b are roots of

x^2 - 6x + P = 0, for some constant P = ab.

Substitute a, then b, and add to yield:

a^2 + b^2 - 6(a + b) + 2P = 0,

from which P = (6*6 - 10)/2 = 13.

Now multiply the quadratic by x:

x^3 - 6x^2 + 13x = 0.

Again, substitute a, then b, and add to yield:

a^3 + b^3 - 6(a^2 + b^2) + 13(a + b) = 0,

from which a^3 + b^3 = 6*10 - 13*6 = -18.

For f(n) = a^n + b^n, where n is a positive integer, this approach can be used to set up a recurrence relation:

f(n+2) = 6f(n+1) - 13f(n).

A similar method can be used to solve the original problem.

Nick


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Graham C
Member since Feb-5-03
Mar-17-04, 03:21 PM (EST)
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6. "RE: Sum of fourth powers"
In response to message #5
 
   The tip was enough. I'd solved the two-variable case by solving the quadratic for a and b and couldn't see any way of solving your problem without solving the cubic, which is not something I care to do unless I have to.

We have
(i) a plus b plus c = 6
(ii) a^2 plus b^2 plus c^2 = 8
(iii) a^3 plus b^3 plus c^3 = 5

a, b and c must be the roots of the equation
(x-a)*(x-b)*(x-c) = 0, or
x^3 - x^2*(a plus b plus c) plus x*(ab plus ac plus bc) - abc

from (i) squared - (ii), ab plus ac plus bc = 14 so we have
x^3 - 6*x^2 plus 14*x - abc = 0

Substituting the roots a for x, b for x and c for x and adding the three together therefore gives

5 - 6*8 plus 14*6 = 3*abc and abc = 15

Multiplying the cubic by x gives
x^4 - 6*x^3 plus 14*x^2 - 15*x = 0

and the same substitution process gives

(a^4 plus b^4 plus c^4) - 36 plus 112 - 75 = 0
(a^4 plus b^4 plus c^4) = -1

which I trust is the answer.

The recurrence relation is now
f(n) = 6*f(n-1) - 14*f(n-2) plus 15*(n-3)



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NIH
Member since Jan-21-04
Mar-17-04, 10:12 PM (EST)
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7. "RE: Sum of fourth powers"
In response to message #6
 
   You have an arithmetic slip, in that abc = 41/3. Then, 6 times the sum of the cubes is 30, rather than 36. But that is the method I was suggesting.

Nick


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Graham C
Member since Feb-5-03
Mar-18-04, 11:09 AM (EST)
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8. "RE: Sum of fourth powers"
In response to message #7
 
   Misquoting:

To make one arithmetic error is understandable. To make two smacks of carelessness.

More to the point, if you don't actually calculate the values of a, b and c, there isn't any simple way of checking the answer, is there? (Simple = other than by checking each step.)


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