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

[jeedoubts]

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?

 

[radou]

Unless I'm missing something, the problem you posted isn't consistent - what do your numbers x, y, z, w satisfy?

 

[jeedoubts]

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

 

[jeedoubts]

please help

 

[Mark44]

Edit: Add "= 1" to make an equation below.

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?

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.

 

[D H]

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.