Prove x2+y2+z2+w2=36 for Real Numbers x, y, z, w

AI Thread Summary
To prove that x² + y² + z² + w² = 36 given the equation (x²/(n²-1)) + (y²/(n²-32)) + (z²/(n²-52)) + (w²/(n²-72)) = 1 for n = 2, 4, 6, and 8, start by substituting these values of n into the equation. This will yield four separate equations involving the variables x, y, z, and w. Solving these equations will provide the necessary relationships between the variables. The approach is straightforward and involves algebraic manipulation to derive the required result. The solution hinges on the consistency of the equations generated from the substitutions.
jeedoubts
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Homework Statement


if the real numbers x,y,z,w satisfy (x2/(n2-1))+(y2/(n2-32))+(z2/(n2-52))+(w2/(n2-72)) for n=2,4,6,8 then prove
x2+y2+z2+w2=36

Homework Equations


The Attempt at a Solution


unable to think of anything?:confused:
 
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Unless I'm missing something, the problem you posted isn't consistent - what do your numbers x, y, z, w satisfy?
 
radou said:
Unless I'm missing something, the problem you posted isn't consistent - what do your numbers x, y, z, w satisfy?


sorry the exact equation is as follows
[(x2/(n2-1))+(y2/(n2-32))+(z2/(n2-52))+(w2/(n2-72))]=1
 
please help:confused::confused:
 
Edit: Add "= 1" to make an equation below.
jeedoubts said:

Homework Statement


if the real numbers x,y,z,w satisfy (x2/(n2-1))+(y2/(n2-32))+(z2/(n2-52))+(w2/(n2-72)) = 1 for n=2,4,6,8 then prove
x2+y2+z2+w2=36




Homework Equations





The Attempt at a Solution


unable to think of anything?:confused:
You're unable to think of anything? The most obvious starting point is substituting n = 2, n = 4, n = 6, and n = 8, and seeing what you get.
 
Mark44 said:
Edit: Add "= 1" to make an equation below.
You're unable to think of anything? The most obvious starting point is substituting n = 2, n = 4, n = 6, and n = 8, and seeing what you get.
That will give you four different equations in four unknowns -- in other words, exactly what is needed to solve the problem.
 

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