Prove that x=y=z in the equations
$\displaystyle x+y^2+z^4=0$
$\displaystyle y+z^2+x^4=0$
$\displaystyle z+x^2+y^4=0$
where x, y, and z are real numbers.
x=-(y^2+z^4)
y=-(z^2+x^4)
z=-(x^2+y^4)
let us consider that x=y=z and let us assume a real value which is positive.
clearly any real number raised to even powers like 2 or 4 must give positive values
as seen from the equation a positive value cannot be equal to a negative value. so the question is wrong.
now let us assume a real negative value.
x=y=z=-a
clearly LHS of the equations are negative.
-a raised to powers of 2 or 4 will produce positive values which are much larger or smaller (in case of negative fractions) than -a; then
due to the negative sign outside bracket RHS become negative values which are much smaller or larger (in case of negative fractions) than -a
Hence we draw conclusions that the equations are wrong.
HELLO FLYINGSQUIRREL
first of all i thank you
1-X-X^3=0 has one real root which comes to be NEARLY equal to 0.68
so x=y=z~0.68 is a solution to the equations
but it is not a strong enough proof as it is a backdoor approach
please reply
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