# Are polynomials of the form : $f_n= x^n+x^{n-1}+\cdots+x^{k+1}+ax^k+ax^{k-1}+\cdots+a$ irreducible over $\mathbb{Z}$?

Is it true that polynomials of the form :

$f_n= x^n+x^{n-1}+\cdots+x^{k+1}+ax^k+ax^{k-1}+\cdots+a$

where $\gcd(n+1,k+1)=1$ , $a\in \mathbb{Z^{+}}$ , $a$ is odd number , $a>1$, and $a_1\neq 1$

are irreducible over the ring of integers $\mathbb{Z}$?

Eisenstein's criterion , Cohn's criterion , and Perron's criterion cannot be applied to the polynomials of this form.

Example :

The polynomial $x^4+x^3+x^2+3x+3$ is irreducible over the integers but none of the criteria above can be applied on this polynomial.

EDIT :

Note that general form for $f_n$ is : $f_n=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ , so condition $a_1 \neq 1$ is equivalent to the condition $k \geq 1$ . Also polynomial can be rewritten into form :

$$f_n=\frac{x^{n+1}+(a-1)x^{k+1}-a}{x-1}$$

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I ran some code to see if mathematica could factor any of these over the integers for generic $a$. It seems like the statement holds for such polynomials of degree up to 150. –  Aleks Vlasev Nov 17 '11 at 9:29
@AleksVlasev,so, you haven't found any counterexample ? –  pedja Nov 17 '11 at 9:33
No, there does not seem to be a counter-example for these. I also ran the code with explicit $a$'s for $a = 2t+1$, $t = 1, 2,3,\dots,50$ and $n \leq 50$. Still nothing. –  Aleks Vlasev Nov 17 '11 at 9:47
@AleksVlasev,If you want to answer the question I think that's ok to post the code as an answer.. –  pedja Nov 17 '11 at 10:03
Could you provide some context for this problem? –  Aleks Vlasev Nov 17 '11 at 20:41

An excellent question!

EDIT: I misread the question where it stated that $a_1 \neq 1$. In the case where $a_1 = 1$ there are infinitely many counter-examples for every odd degree $\ge 3$.

Here is the code I used. It is far far from perfect, so feel free to try it out and improve on it.

f[n_, k_, a_, z_] := a Sum[z^i, {i, 0, k}] + Sum[z^i, {i, k + 1, n}]

falselist = {};

For[t = 1, t <= 50, t++,
list = Table[If[GCD[n + 1, k + 1] == 1, Expand[f[n, k, 2 t + 1, z]], 0], {n, 1, 50}, {k, 1, n - 1}];
list = Select[Flatten[list], SameQ[#, 0] == False &];
new = Table[{Factor[list[[k]]], SameQ[list[[k]], Factor[list[[k]]]]}, {k, 1, Length[list]}];
For[k = 1, k <= Length[new], k++,
If[new[[k, 2]] == False,
falselist = Append[falselist, {Expand[new[[k, 1]]], new[[k, 1]]}]]
]
]

falselist


What the code does is the following. It first creates a list of all the relevant polynomials of degree up to and including 50. To do this, the code just runs over all possibilities and checks which ones satisfy the gcd requirement. Then the code picks the non-zero entries (because we don't want any Null's).

Then the code creates a table (new) where each entry is the pair $f_n$ and whether the factoring fails (so SameQ would give the answer "True"). In the end, the for-loop picks the polynomials that do factor and factors them.

The outside for-loop goes over the odd $a$'s up to 101.

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thank you for your interesting in my question but it seems that you didn't understand me well. There is condition in statement $a_1 \neq 1$ and all of your "counterexamples" have coefficient $a_1=1$ –  pedja Nov 17 '11 at 19:45
$f_n=a_nx^n+a_{n-1}x^{n-1}+....+a_1x+a_0$ –  pedja Nov 17 '11 at 19:46
Oh, I see. I'm sorry for misreading the question. So, the condition is $k\ge 1$. –  Aleks Vlasev Nov 17 '11 at 19:49
I changed my answer and the code as well to take into account that $k\ge 1$. –  Aleks Vlasev Nov 17 '11 at 19:52
yes with $a_1\neq 1 \Rightarrow k\geq 1$ –  pedja Nov 17 '11 at 19:54

Aleks idea produces lots of counterexamples in case $k=0$.

let $P(x)=x^n+x^{n-1}+..+x$, with $n$ odd. Pick any negtive integer $a$ so that $P(a) <0$.

Then $P(x)-P(a)$ is reductible over $Z$.

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The case $a=2$ is solved in the following post on my blog: http://mathproblems123.wordpress.com/2009/11/02/irreductible-polynomial/ It is a part of Miklos Schweitzer 2009 contest. A generalization can be found in the second part of the post, generalization which I proposed for the Romanian Team Selection Tests for IMO in 2010.

The polynomial you present clearly satisfies the conditions of the following property: http://mathproblems123.wordpress.com/2009/11/09/position-of-roots/, which shows that the roots of your polynomial have modulus strictly greater than 1 or your polynomial has as a root a root of unity of some order. This might help you derive some properties which prove irreducibility, using roots.

When $a$ is prime, the method in the first post can be applied and your polynomial is irreducible. When $a$ is composite, things might get trickier.

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Supposing that there were two disisors of $f_n$, with endings $(...gx^{2}+cx+d)(...hx^{2}+ex+f)$. Then a=df and a=de+fc and a=hd+gf+2ce. The first two equations are only possible if d and f share at least one common factor p which is not a factor of c or e. But then by the third equation p|2ce, and p must be odd since p|a and a is odd.

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Why $2ce$? Also, isn't $d=2$, $f=3$, $e=3$, $c=0$ a counterexample to your third sentence? –  user7530 Dec 1 '11 at 10:14