**False.**

Proof:

First, let us prove a short lemma:

For any

:

if and only if

and

have the same powers of of

and same coefficients.

The proof is simple:

Let

be the absolute value of the most negative power of

in

(

if there are no negative powers). For example, when

and

we have

. Thus

, and therefore the powers and coefficients of

and the same (since we assumed

for all

). The result follows.

Back to the original problem:

Let

and

. As before, let

be the absolute value of the most negative power of

in

, and

if no negative powers exist. Thus, the most negative power in

for any

is no lower than

. Therefore, the most negative power of

in the set

is

. From the lemma above and from the fact that the powers of

in Laurent polynomials are not bounded, we conclude that there exist a Laurent polynomial

for which

, so

for any choose of

and

. In other words, it is

**false** that there exist an inetger

and

such that

.