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MIT PRIMES/Art of Problem Solving

CROWDMATH 2024: Where Does the Goldbach Conjecture Hold?

Where Does the Goldbach Conjecture Hold? Polymath project forum
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Where Does the Goldbach Conjecture Hold? Polymath project forum
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exercise 1.3
aeemc2   2
N Nov 3, 2024 by aeemc2
One of the irreducible criteria is well-known for polynomials with integers coefficients and also holds naturally for polynomials of $\mathbb{N}_0[x]$ is Eisenstein's irreducible criterion that states
Criterion 1 (Eisenstein's irreducible criterion).
Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 \in \mathbb{N}_0[x] $. If there exists a prime number $p$ satisfying the following three conditions:

- $p$ divides each $a_i$ for $0 \leq i < n$.
-$p$ does not divide $a_n$, and
-$p^2$ does not divide $a_0$.

Then the polynomial $p(x)$ is irreducible.

We have another criterion, which is even easier to check.

Criterion 2: Let $p(x) \in  \mathbb{N}_0[x]$. Suppose that $p(0) \neq 0$ and $p(1)$ is a prime number, then $p(x)$ is irreducible.

Proof
Indeed, suppose that $p(x)=q(x)r(x)$ for some $q(x), r(x) \in    \mathbb{N}_0[x]$. Note that $p(1)=q(1)r(1)$, and since $p(1)$ is prime, we know that either $q(1)=1$ or $r(1)=1$. WLOG, assume that $r(1)=1$, then $r(x)=x^n$ for some $n \in \mathbb{N}_0$. Note that if $n > 0$, then $r(0)=0$, and then $p(0)=0$, which contradicts the hypothesis. Therefore $r(x)=1$, and this implies that $p(x)$ is irreducible.

2 replies
aeemc2
Nov 1, 2024
aeemc2
Nov 3, 2024
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