Polynomial

[Euler_India]

Let f(x) be a polynomial with real coefficients and degree greater or equal to 1.Show that for every real number c>o,there is a positive integer n satisfying the following condition: If the polynomial P(x) of degree greater or equal to n with real coefficients and leading coefficient equal to 1. Then the number of integers x for which Modulus f(P(x)) \leq c is not
greater than the degree of P(x).

[Myth]

Hello, Euler_India!

I have long proof.

Let c>0.

If |f(y)|<c then |y|<K_c where K_c is some finite number. Let C=max(2,K_c).

Now we will prove following statement.
There is a positive integer n satisfying the following condition: If the polynomial P(x) of degree greater or equal to n with real coefficients and leading coefficient equal to 1 then the number of integers x for which |P(x)|<C is not greater than the degree of P(x).

Let n₀=20*C.

Consider P(x), deg(P(x))=n>n₀. If P(x) has non-real root a then conjugate complex number a' also root of P(x). Replacing a and a' with Re(a) we obtain new polynomial P'(x) and |P'(x)|<|P(x)| ==> set {x | |P(x)|<C} is smaller then {x | |P'(x)|<C}. So we can WLOG suppose P(x) has real roots only, denote them a₁,...,a_n, P(x)=(x-a₁)...(x-a_n).

Let t₁, ..., t_n+1 - distinct integers. Suppose |P(t_i|<C for all i=1,2,...,n+1.
It is clear there is t_k such that |t_k-a_i| \geq 1/2 for all i=1,2,...,n. Suppose WLOG |t_k-a_i|<=2C, i=1,2,...,s and |t_k-a_i|>2C, i=s+1,...,n. We have C>|P(t_k)|>(1/2)^s(2C)^n-s ==> C^n-s-1 < 2^2s-n ==> n-s-1 < 2s-n ==> s>=2n/3. Segment [t_k-2C-2,t_k+2C+2] contains 4C+5 integer numbers ==> there n-4C-5 of t_i (i=1,2,...,n+1) outside [t_k-2C+2,t_k+2C+2] and only n/3 of a_i (i=1,...,n) ==> within such t_i exists t_p such that |t_p-a_i|>=1 for all i=(s+1),...,n ==> C>|P(t_p)|>2^s>=2^2n/3>C !!!!

P.S. Some inequalities denote order of values only (for example, s>=2n/3) and haven't exact meaning.