Method to find if a polynomial has irrational roots?

Question

Question

Let's say I define a polynomial $P(x)$ whose roots $\alpha_i$ and degree $> 1$. We also add constraint that the coefficient of the highest power of $P(x)$ is $1$ and all other coefficients are integers.

when $P(0) \neq 0$. Then:

$$ \frac{P(|P(0)|)}{P(0)} = \prod_{i=1}^{\deg P(x)} (1 - \frac{P(0)}{\alpha_i}) $$

where |x| is the modulus of x. Let, the number of prime factors of ${P(|P(0)|)}/{P(0)}$ be $z$. Then if,

$$ \deg P(x) > z$$

and $P(2) \neq 0$ this implies $P(x)$ has some irrational roots.

Example

$$P(x) = x^4 -x^3 + 3x^2 + 3$$

thus,

$$\frac{P(|P(0)|)}{P(0)} = 7 \times 2 \times 2$$

Hence, the number of primes are $3$ but $\deg P(x) = 4 > 3$. We conclude, $P(x)$ has irrational roots.

Question

Is it possible to quantify how often a polynomial with some irrational roots of the form $P(x)$ satisfies $ \deg P(x) > z$?

Answer 1

If $P$ of degree $n$ and with $P(0)\ne 0$ has only rational roots, then these must in fact be (non-zero) integers and $P(0)$ is (up to sign) their product. Thus $$\tag1 \frac{P(|P(0)|)}{P(0)}=\frac{\prod_{i=1}^n(|P(0)|-\alpha_i)}{\prod_{i=1}^n(0-\alpha_i)}=\prod_{i=1}^n\left(1-\frac{|P(0)|}{\alpha_i}\right)$$ is the product of $n$ integers. As $P(0)\ne 0$, none of the factors is $=1$. If additionally $P(|P(0)|)\ne 0$, none of the factors is $=0$.

Case 1. If none of the factors in $(1)$ is $=-1$, then all $n$ factors are integers $\notin\{-1,0,1\}$ and hence each contributes at least one prime factor. So in this case $z\ge n$, as desired.

Case 2. Now suppose that at least one of the factors in $(1)$ is $=-1$, which happens precisely when $|P(0)|=2\alpha_i$ for some $i$. Note that in this case, the product of the other $n-1$ roots is $\pm2$, i.e., exactly one of them is $\pm2$ (but by the additional condition that $P(2)\ne 0$, it must be $-2$) and the rest are $\pm1$. In other words, $$ P(X)=(X-\alpha)(X+ 2)(X-1)^{n_1}(X+1)^{n_2}$$ for some positive integer $\alpha$ and with $n_1+n_2=n-2$. Explicitly, this makes $|P(0)|=2\alpha$ and $$\tag2\frac{P(|P(0)|)}{P(0)}=\frac{P(2\alpha)}{P(0)}=(-1)\cdot (\alpha+1)\cdot (1+2\alpha)^{n_1}(1-2\alpha)^{n_2}$$

• If $\alpha=1$, this becomes $(-1)^{n_2+1}\cdot 2\cdot 3^{n_1}$ and has $z=n_1+1$ prime factors. As this makes $z<n$, we obtain counterexamples to the claim! For a concrete counterexample, consider $$ P(X)=(X-1)(X+2)=X^2+X-2,$$ for which $\frac{P(|P(0)|}{P(0)}=\frac{P(2)}{-2}=-2$ has only one prime factor, but has only rational roots.

• We cannot have $\alpha=2$ as that would make $P(2)=0$.

• If $\alpha\ge 3$, each factor in $(2)$ except the first is $\notin\{-1,0,1\}$, hence the only way to have $z<n$ is when all of these are $\pm$ a prime. In particular $(2\alpha-1,2\alpha+1)$ must be a pair of twin primes. As $\alpha=2$ is excluded, these twin primes must be of the form $6k\pm1$, i.e., we must have $\alpha=3k$ and $\alpha+1=3k+1$ also prime. This happens when $a\in\{6, 30,36,96,156,210,330,546,576,660,726,810,936,966,\ldots\}$ and so gives rise to many more counterexamples. Here's a concrete les trivial one: $$ P(X)= X^4 - 94X^3 - 193X^2 + 94X + 192$$ has $P(0)=192$, $P(192)=686536512$, $\frac{P(192)}{P(0)}= 97\cdot 191\cdot 193$, so $z=3<n=4$.