Math Is Fun Forum

jazzpiano

Guest

Difficult equations

Solve in real numbers:

x^5 = 5y^3 - 4z
y^5 = 5z^3 - 4x
z^5 = 5x^3 - 4y

I konw that solutions may be x=y=z = -2,-1,0,1,2 , but I can't proof that's it's the only solution.

Please help!

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Difficult equations

You can see that the three equations taken together are symmetrical in x, y, z – meaning if you permute x, y, z in a cyclic fashion, you get the same three equations (only in a different order). Also, all the powers involved are odd. Hence x, y, z must all be equal, because if any pair of them were to be different, the symmetry would be broken.

So substitute x = y = z into any of the equation and solve the easy quintic equation.

NB: If you have a system of cyclically symmetrical equations involving even powers, you might have ±x = ±y = ±z.

Last edited by JaneFairfax (2007-09-09 22:26:25)

kmadhyan

Member

Registered: 2007-08-30

Posts: 4

Re: Difficult equations

hi , The solution of this is simple
x^5 = 5y^3 - 4z
y^5 = 5z^3 - 4x
z^5 = 5x^3 - 4y

Since all three equations are rotational
like
x=y+z
y=x+z
z=x+y
That is they rotate from preceding equation
This implies
x=y=z

So just place y & z with x from any of the equation
x^5 = 5x^3 - 4x
=>x^5-5x^3 + 4x=0
=>x(x^4-5x^2 + 4)
=>x((x^2 -4)(x^2 -1))
=>Hence now it is easy I suppose
x=-2,-1,0,1,2 =y=z

You can take any value at the time and can verify

---

IT is maths which has brought humanity from Ice Age to Present

Aaron123

Guest

Re: Difficult equations

Is there any other solution to show, that x=y=z should be equal?