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NIH
Member since Jan2104

Mar0804, 06:19 PM (EST) 

"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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alexb
Charter Member
1230 posts 
Mar0804, 09:39 PM (EST) 

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 x^{2} + y^{2} + z^{2} = 4 x^{3} + y^{3} + z^{3} = 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

Mar0904, 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*8^26*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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Graham C
Member since Feb503

Mar1204, 11:03 PM (EST) 

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 Jan2104

Mar1504, 07:05 PM (EST) 

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 Feb503

Mar1704, 03:21 PM (EST) 

6. "RE: Sum of fourth powers"
In response to message #5

The tip was enough. I'd solved the twovariable 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 (xa)*(xb)*(xc) = 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(n1)  14*f(n2) plus 15*(n3)


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