on polynomial roots

let f() have a real root c₀.
then c₁=c₀²+c₀+1 is also a root.
check c1≥3/4.
define c_i=c_i-1²+c_i-1+1
then all ci 's are real roots of f. also c2>1 implying ci>c()i-1 for all i>2..
but f can have only finitely many roots. so f doesn't have any real root.

gr8 work ith_power..

For those who are still wrestling with the problem ...remember that odd power polynomials must have at least one root...(try why?) ...so just show that its impossible for f(x) to have any real root..

Good work guys..Its time for me to increase the difficulty of questions :)

In fact, to prove rigorously that f cannot have any real root, you will have to prove that it is not possible to make a ring of roots,

r_o, r₁,...., r_n satisfying

r₁ = r₀²+r₀+1

r₂ = r₁²+r₁+1
.
.
.
r₀ = r_n²+r_n+1

as this give rise to the absurdity that both

r₀<r₁<r₂<...<r_n and

r₀> r_n hold simultaneously

Which means that f(x) has all roots non-real

Now, if we add the condition that f(x) has all coefficients real, we can state that f(x) is of even power.

(If all roots are non-real then they occur in conjugate pairs) and so there are an even number of roots. which means the degree of f is even)