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I want to prove that $f(x)=10x^4-18x^3+4x^2+7x+16 \in \mathbb{Z}[x]$ is irreducible in $\mathbb{Q}[x]$.

If $f(x)$ can be analyzed in a product of two non-constant polynomials with coefficients $\in \mathbb{Q}$,then $f(x)$ can be analyzed in a product of two non-constant polynomials with coefficients $\in \mathbb{Z} $.

One case is: $f(x)=(ax^2+bx+c) \cdot (a'x^2+b'x+c')$. So,to check if there are such polynomials,I have to solve the system:

$$\left\{\begin{matrix} x^4: & 10=aa' \\ x^3: & -18=ab'+ba' \\ x^2: & 4=ac'+a'c+bb' \\ x: & 7=bc'+cb' \\ \text{ constant: } & 16=cc' \end{matrix}\right.$$

So,at the beggining,I suppose that $a=10,a'=1$ and then $a=5,a'=2$.

Then,do I also have to solve the system, with $a'=10,a=1$ and $a'=5,a=2$ or is it the same??

evinda
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    By symmetry it is the same. The idea is that $(ax^2 + bx + c)(a'x^2 + b'x + c') = (a'x^2 + b'x + c')(ax^2 + bx + c)$ – user88595 May 07 '14 at 13:34
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    Note that $f(1)=19$ is prime, so if it factors then one of the quadratics must evaluate to $\pm1$ at $x=1$. – Gerry Myerson May 07 '14 at 13:36
  • @user88595 so there are only two possible values for $a$,$10,5$ and respectively for $a'$, $1,2$,right? – evinda May 07 '14 at 13:37
  • @GerryMyerson I haven't understood it..could you explain it further to me? – evinda May 07 '14 at 13:38
  • What don't you understand? Do you not understand the notation $f(1)$? Do you not understand why $f(1)=19$? Do you not understand what a prime number is? Do you not understand why 19 is prime? Help me out here. – Gerry Myerson May 07 '14 at 13:40
  • @evinda : without loss of generality yes, otherwise you are simply repeating yourself. As for Gerry's comment, what he meant is that $f(x) = f_1(x)f_2(x)$ for some lower degree polynomials. Since $f(1)$ is prime then either $f_1(x)$ or $f_2(x)$ has to be $\pm1$ by prime properties. – user88595 May 07 '14 at 13:45
  • A ok I understood it now..But how can I use it? – evinda May 07 '14 at 13:46
  • @evinda : that means $a\cdot 1^2 + b\cdot 1 + c = \pm 1$ which could help find an expression, or prove there aren't any. – user88595 May 07 '14 at 18:26
  • @GerryMyerson I showed a way to push through using your observation. – Calvin Lin May 07 '14 at 20:49

3 Answers3

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As a first step have fun checking out that the polynomial has no rational zeros. Leaving that to you [insert evil grin].

I am considering the possibility of $f(x)$ having two quadratic factors with integer coefficients. Let us project from the ring $\Bbb{Z}[x]$ to $\Bbb{Z}_2[x]$ by reducing all the coefficients modulo two. If $f(x)=g(x)h(x)$ for two quadratics $g(x),h(x)\in\Bbb{Z}[x]$, then modulo two we have $$ \overline{g}(x)\overline{h}(x)=\overline{f}(x)=x. $$ Because the ring $\Bbb{Z}_2[x]$ is a unique factorization domain, this means that one of $\overline{g}(x)$ and $\overline{h}(x)$ is $=x$ and the other $=1$.

In particular this implies that the leading coefficients of both $g(x)$ and $h(x)$ must be even for otherwise that factor would remain a quadratic after reduction modulo two as would then the product.

But this would imply that the leading coefficient of $f(x)=g(x)h(x)$ would be divisible by four, which is manifestly not the case. Thus such a factorization does not exist.

Q.E.D.

Jyrki Lahtonen
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Here is another approach, that is nice, but is actually more tedious than the rest.

Suppose not. Let $ f(x) = g(x) h(x)$ where $g$ and $h$ are integer quadratic polynomials.

Observe that $ f(-15), f(-7), f(-5), f(-3), f(-1), f(1), f(3), f(7), f(25)$ are all prime.

WLOG, $g(x) = \{1, -1\}$ for 5 of those values. WLOG $g(x) = 1$ for 3 of those values. Hence, $g(x) = 1$ is a constant.

Note: This is actually more tedious, because you need to check for primality. I let the computer do that. It is likely easier to try and solve the Diophantine equation.

Calvin Lin
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The polynomial $f(x)=10x^4-18x^3+4x^2+7x+16$ is irreducible iff $g(t)=f(t+1)$ is irreducible. We have $$g(t)=10t^4+22t^3+10t^2+t+19.$$ There being no linear factors, the only factorization possible would be into $$g(t)=(at^2+bt+c)(pt^2+qt+r).$$ Then $ap=10$ and without loss one may assume either $a=10,p=1$ or else $a=5,p=2.$ [possible negative choices could be removed since $-1$ could be factored out of all coefficients, and we may assume it is the first quadratic which begins with coefficient $10$ or $5$.

Now we also have that $cr=19,$ a prime. For each possible choice for the four coefficients $a,p,c,r$ we may use the $t^2$ coefficient of $10$, so that $$ar+bq+cp=10,$$ and we may insert the values of $a,p,c,r$ and obtain the value of $bp$. In each case $bp$ is either a prime or a product of two primes, and one may check that the $t$ coefficient, which should be $1$, does not match. There are eight possible choices for the values of $(a,p,c,r)$ as follows, where the number after the four is the value of $bp.$

$(10,1,19,1;\ -19)$

$(10,1,-19,-1;\ 39)$

$(10,1,1,19;\ -181)$

$(10,1,-1,-19;\ 201)$

and four more starting out $(5,2,...)$.

In each case we can factor the value obtained for $bq$ and give assigned values to $b,q$ accordingly. By only checking the $t^1$ coefficient, which gives $br+cq=1$, it turns out that none of the possibilities gives this last equation. For example in the second case, where $bq=39=3\cdot 13$ we try $(b,q)=(39,1),(1,39),(13,3),(3,13)$ and none give the right value of $1$ to $br+cq.$ [In checking one may assume that say $b>0$ for setting up the cases, since a simultaneous sign change on $b,q$ only changes the sign of the quantity $br+cq$, and so we only need see if this ever comes out $\pm 1$.

This is admittedly a tedious computation, but at least it shows irreducibility.

coffeemath
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