# An irreducibility problem.

Here is a problem from one of the Harvard quals question paper(spring 2011 I think) which I could not figure it out for a long time.

Prove that for any positive integer $a$, the polynomial $f(x)=x^6+3ax^4+3x^3+3ax^2+1$ is irreducible.

I tried to use Eisenstein criterion but I couldn't find a suitable transformation to $f$ so that Eisenstein can be applied and I don't know any other criterion other than this( If we need some other criterion to solve this please mention it).

Thanks.

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If $a$ is even, then $f$ is irreducible modulo $2$, so it must be irreducible over $\mathbf{Z}$ as well. Odd $a$ leaves open the possibility that $f$ is a product of quadratic and a quartic. Modulo $3$ doesn't give anything more. Also $f$ is palindromic, so as taking the reciprocal won't switch the two factor, they must be palindromic as well. – Jyrki Lahtonen Apr 26 '12 at 17:27
One more attempt at rephrasing. For $p(x)$ of degree $n$, write $\tilde{p}(x)=x^{-n}p(1/x)$. If $f(x)=g(x)h(x)$ is a factorization over the integers, and $f=\tilde{f}$, then $$f(x)=\tilde{f}(x)=\tilde{g}(x)\tilde{h}(x)$$ is another. In the remaining case here we have $\deg g=2$, $\deg h=4$, so we can conclude $\tilde{g}=\pm g$ and $\tilde{h}=\pm h$, because $\pm1$ are the only units. The known modulo 2 reduction of a putative factorization tells us that the linear term of $g(x)$ and the quadratic term of $h(x)$ are non-zero, which rules out the minus signs. – Jyrki Lahtonen Apr 27 '12 at 3:07

Idea:

If $f(x)= P(x)Q(x)$, with $P,Q$ monic, then in $Z_3[x]$ you have

$P(x)Q(x)= x^6+1=(x^2+1)^3$.

$x^2+1$ is irreducible over $Z_3[x]$, thus the only possibility (up to renaming) is $P(x) \equiv x^2+1 \mod 3$ and $Q(x) \equiv (x^2+1)^2 \mod 3$.

I think the problem should follow easily from here, express $P(x)=x^2+1+3P_1(x)$, $Q(X)=(X^2+1)^2+3Q_1(X)$, with $\deg(P_1) \leq 1, \deg(Q_1) \leq 3$ and plug it back into the first equation....

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why is $degP_1$ less than 1? – Dinesh Apr 26 '12 at 17:34
because $P(x)$ and $Q(X)$ are monic. Degree of $P_1$ cannot be 2, because the leading coefficient of $P$ would not be $1$, and cannot be more than 2, because $\deg(Q)$ is at least 4.... – N. S. Apr 26 '12 at 17:37
P.S. I think that Jyrki's observation that the factors need to be palindromic easily finish the proof... The only way $P$ is palindromic is if $P_1 =3bx$. In order for $Q$ to be palindromic, we have $Q_2=3cx^3+3dx^2 +3cx$. – N. S. Apr 26 '12 at 17:39
Thanks that solves the problem i guess. – Dinesh Apr 26 '12 at 17:46
The coefficients are the same if read backwards. A polynomial of degree $n$ is palindromic if and only if $P(x)=x^nP(\frac{1}{x})$. – N. S. Apr 26 '12 at 18:13

I think it should be sufficient to consider the reduction mod 2. If $a$ is even, then the polynomial reduces to $x^6+x^3+1$. If $a$ is odd, then the polynomial being considered reduces to $x^6+x^4+x^3+x^2+1$. Clearly, neither of these polynomials have constant factors. If they can be reduced then it must be as a product of an irreducible quadratic and an irreducible quartic, two irreducible cubics, or as a product of three irreducible quadratics. Notice that in $\mathbb{Z}/(2)[x]$, the only irreducible quadratic is $x^2+x+1$. But neither polynomial has $x^2+x+1$ as a possible factor. Thus, if the two polynomials are reducible modulo 2, then they must be a product of two irreducible cubics. There are only two irreducible cubics in $\mathbb{Z}/(2)[x]$: $x^3+x^2+1$ and $x^3+x+1$. A quick computation shows this to be impossible.

Since the polynomial is irreducible modulo 2 for every $a$, the polynomial $x^6+3ax^4+3x^3+3ax^2+1$ is irreducible for every $a$.

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Thats quite an elegant solution! thanks! – Dinesh Apr 26 '12 at 18:00
Unfortunately this is incorrect. Modulo 2 we have $$x^6+x^4+x^3+x^2+1=(x^2+x+1)(x^4+x^3+x^2+x+1).$$ – Jyrki Lahtonen Apr 26 '12 at 18:07
Jyrki Lahtonen hmmm! :( – Dinesh Apr 26 '12 at 18:16
You are absolutely correct; I'm so sorry for my misleading comments. I messed up part of my calculations and thought I had redone them all, unfortunately that wasn't the case. Once again, sorry. – chris Apr 26 '12 at 18:38

Here is what I tried. This does not cover the case of quadratic-times-quartic.

First, we can use synthetic division to determine that there are no linear factors. For example, $f(-1) = 6a -1 = 0 \Rightarrow a \notin \mathbb{Z}^+$

Then I considered $$f(x) = (x^3 + bx + c + 1)(x^3 -bx^2 - cx + 1)$$

The justification of these coefficients is the absense of the fifth- and first-degree terms in $f$. From here, we see (if my calculations are correct!) that $3a = -b^2$, impossible for our values of $a$.

Surely there is a more theoretic/less brute force approach, but I thought this was worth mentioning.

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