How one can reformulate the sentence: $T_{n}(x)/x$ is irreducible over the integers The motivation to this question can be found in: Chebyshev Polynomials and Primality Testing
My question is: 
How one can reformulate the sentence: 
$T_{n}(x)/x$ is irreducible over the integers
into an equivalent statement in the form of an inequality or several inequalities.
Here $T_{n}(x)$ is the Chebyshev polynomial of the first kind.
 A: What you want doesn't exist (otherwise factoring integers would be easy).
An unhelpful answer:
Let $((a \pmod{b}\,))$ be the least non-negative member of the equivalence class $a \pmod{b}$.  $T_n(x)/x$ is irreducible over $\Bbb{Z}[x]$ if and only if the inequalities in the following set are satisfied: 
$$  \left\{ ((n \pmod{d}\,))>0 \mid 2 \leq d \leq n-1  \right\}  \text{.}  $$
We can make an even larger set to check by checking that the remainder on polynomial division of $T_n(x)/x$ by each polynomial of degree less than $n-1$ has degree $<1$ and constant coefficient $>0$.
Another unhelpful answer:
Say we wish to show $T_\nu(\chi)/\chi$ is irreducible over $\Bbb{Z}[\chi]$.
Using the Jones, Sato, Wada, and Wiens example,
Do there exist $a,b,c,d,\dots,z \geq 0$ such that 
$$  \nu = (k+2)\{1 - [wz+h+j-q]^2 - [(gk+2g+k+1)(h+j)+h-z]^2 - [2n+p+q+z-e]^2 - [16(k+1)^3(k+2)(n+1)^2+1-f^2]^2 - [e^3(e+2)(a+1)^2+1-o^2]^2 - [(a^2-1)y^2+1-x^2]^2 - [16r^2y^4(a^2-1)+1-u^2]^2 - [((a+u^2(u^2-a))^2 -1)(n+4dy)^2 + 1 - (x+cu)^2]^2 - [n+l+v-y]^2 - [(a^2-1)l^2+1-m^2]^2 - [ai+k+1-l-i]^2 - [p+l(a-n-1)+b(2an+2a-n^2-2n-2)-m]^2 - [q+y(a-p-1)+s(2ap+2a-p^2-2p-2)-x]^2 - [z+pl(a-p)+t(2ap-p^2-1)-pm]^2\}$$
If so, $T_\nu(\chi)/\chi$ is reducible.  If not, $T_\nu(\chi)/\chi$ is irreducible.
