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I do not have an idea on how to approach this problem. Similar problems that I found online try to show that there does not exist a linear factor of the given equation in $\mathbb{Q}$. But my question is, how is that sufficient to prove that the polynomial is irreducible? It could also be a product of two quadratic polynomials which do not have a solution in $\mathbb{Q}$. Any help is appreciated!

PseudonymousAJ
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    Modulo $2$ it factors as $(x+1)(x^3+x^2+1)$ with the cubic factor irreducible in $\Bbb{Z}_2[x]$. Do you see why this implies that it cannot be a product of two quadratics over $\Bbb{Z}$? – Jyrki Lahtonen Nov 30 '22 at 08:43
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    Please include the problem also in the text / question (not only in the title). Please share with us the own effort to solve the problem, showing explicit steps. https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question As a hint, there is a result (Gauss) insuring that a (non-trivial) factorization over $\Bbb Q$ with two monic factors already lives over $\Bbb Z$. (The coefficients of the factors are thus all integers, when supposed to be only rational.) Now consider the (supposed to exist) factorization modulo some small primes... – dan_fulea Nov 30 '22 at 10:12
  • "my question is, how is that sufficient to prove that the polynomial is irreducible?" It is not, you indeed need to rule out quadratic factors as well (the source you mention probably only prove irreducibility of cubic or quadratic where ruling out linear factors is sufficient). – Sil Nov 30 '22 at 13:48

2 Answers2

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Write $f(x):=x^4-5x^2+x+1$. Since $f$ is monic, whenever $f=gh$ in $\mathbb{Z}[x]$, we may assume without loss of generality that both $g$ and $h$ are monic. You already know that $\operatorname{deg}(g)=1$ is impossible. Assuming $\operatorname{deg}(g)=\operatorname{deg}(h)=2$, we have $$x^4-5x^2+x+1=(x^2+ax+b)(x^2+px+q)\\=x^4+(a+p)x^3+(q+ap+b)x^2+(aq+bp)x+bq$$ for some integers $a,b,p,q$. Comparing coefficients, either $b=q=1$ or $b=q=-1$, but it is easy to check that both cases lead to contradiction. Hence $f$ is irreducible in $\mathbb{Z}[x]$.

Now use Gauss's lemma to conclude that $f$ is irreducible in $\mathbb{Q}[x]$.

Dilemian
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For the degree 1 factors it's sufficient to show that it has no rational solutions as you said.

If this polynomial were the product of two polynomials of degree $2$, then the coefficients of the degree $2$ and $0$ terms can only be $1$ or $-1$. One of its factors would be $x^2+ax + c$, while the other one $x^2+bx + c$, where $c \in \{ -1,1 \}$. Now it's easy since we have only one real parameter.

We can immediately see that, since in the original polynomial we don't have the degree 3 term, $a=-b$. For the degree 2 term, we have $2c + ab=-5$. Putting together these two facts, we have: $$2c-a^2=-5 \implies a^2=2c+5 \implies a^2=b^2=3 \text{ or } a^2=b^2=7 \ , $$ but none of these two equations has an integer solution, so the original polynomial is irreducible.

SilvioM
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