Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

Please give your ideas/hints to solve this.

Thanks.

share|improve this question
    
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

3 Answers 3

up vote 5 down vote accepted

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....

share|improve this answer
    
why is $degP_1$ less than 1? –  Dinesh Apr 26 '12 at 17:34
2  
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
1  
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 $a$.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.