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In this document on page $3$ I found an interesting polynomial:


Question is whether this polynomial is irreducible over $\mathbf{Q}$ for arbitrary $n \geq 1$ ?

In the document you may find a proof that polynomial is irreducible whenever $n=2p , p \in \mathbf{P}$ and below is my attempt to prove that polynomial isn't irreducible whenever $n$ is odd number greater than $3$. Assuming that my proof is correct my question is :

Is this polynomial irreducible over $\mathbf{Q}$ when $n=2k , k\in \mathbf{Z^+}$ \ $\mathbf{P} $ ?

Lemma 1: For odd $n$, $a^n + b^n = (a+b)(a^{n-1} - a^{n-2} b + \cdots + b^{n-1})$.

Binomial expansion: $(a+b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n-1} b + \cdots + \binom{n}{n} b^n$. $$\begin{align*} f(x) &= \frac{(x+1)^n - x^n-1}{x} = \frac{(x+1)^n - (x^n+1)}{x} \\ &= \frac{(x+1) \cdot (x+1)^{n-1} - (x+1)(x^{n-1}-x^{n-2}+ \cdots + 1)}{x} \\ &= \frac{(x+1) \cdot (x+1)^{n-1} - (x+1)(x^{n-1}-x^{n-2}+ \cdots + 1)}{x} \\ &= \frac{(x+1) \cdot \left[ \Bigl(\binom{n-1}{0}x^{n-1}+ \binom{n-1}{1} x^{n-2} + \cdots + 1 \Bigr) - (x^{n-1}-x^{n-2}+ \cdots + 1) \right]}{x} \\ &= \frac{(x+1) \cdot \left[ \Bigl(\binom{n-1}{0}x^{n-1}+ \binom{n-1}{1} x^{n-2} + \cdots + \binom{n-1}{n-2} x \Bigr) - (x^{n-1}-x^{n-2}+ \cdots - x) \right]}{x} \\ &= (x+1) \cdot \small{\left[ \left(\binom{n-1}{0}x^{n-2}+ \binom{n-1}{1} x^{n-3} + \cdots + \binom{n-1}{n-2} \right) - (x^{n-2}-x^{n-3}+ \cdots - 1) \right]} \end{align*}$$

So, when $n$ is an odd number greater than $1$, $f_n$ has factor $x+1$. Therefore, $f_n$ isn't irreducible whenever $n$ is an odd number greater than $3$.

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About the case when $n$ is odd: $f_n(-1)=0$ hence $x+1$ divides $f_n(x)$, end of the proof. – Did Nov 15 '11 at 8:16
The document says that it is an open problem. – Dinesh Nov 15 '11 at 9:28
It is open because it has not yet been determined for all even $n$. – Cameron Buie May 14 '12 at 21:54
When you ask for $f_n$ to be irreducible, does this include constant factors? It appears that the content of $f_n$ is $1$ unless $n=p^k$ is a prime power, in which case the content is $p$. – A Walker May 9 '13 at 22:18
@A. Walker: You are right. It is only interesting to ask for irreducibility over $\mathbb{Q}$. From $f_4 = 2 (2 x^2 + 3 x + 2)$ I infer that actually this was meant, therefore I've edited the question. – Martin Brandenburg May 27 '13 at 21:25

1 Answer 1

Polynomials are just numbers in base x. All numbers representable in base x which are divisible by base x numbers are divisible for ALL bases. Following from this reasoning, I will substitute 10 for x so that we can work within a more comfortable situation (it might even prove to be obvious and is a good technique for factoring polynomials!).

$f(n) = (11^n - (10^n +1))/10$

All powers of 11 start and end with a digit with a coefficient of 1. Carry-over may break this, but remember that it is a representation and that carry-over will not apply to base x. Division by 10 will shift the numbers over to the right. This means that the final result will consist of the inner terms of x divided by 10. The question then abstractly asks: "are the inner terms of all powers of 11 divisible by any number other than 10".

n = 1 is a trivial identity as the number will have no inner digits and therefore be 0, which is divisible by everything.

For n = 2, the powers of 11 are divisible as the first set of inner digits are even. Granted, there is implied carry-over and so this will not be valid for base x. I.E. even number divisibility doesn't transcend base 10 except for the term n = 2 whereas there is only one inner term.

There is a simple explanation to this issue. You see, all powers of 11 work off of a pyramid. The next terms in the pyramid are equal to the sum of the coefficients of the two terms above it.

This provides 3 basic rules regarding these numbers:

the innermost digit is always even (except when the number of digits is even)

the edge inner terms are equal to the current power of 11

the numbers are parallel with regards to the digit n/2

look at the following pyramid:

              1   1
            1   2   1
          1   3   3   1
        1   4   6   4   1
      1   5  10  10   5   1
    1   6  15  20  15   6   1
  1   7  21  35  35  21   7   1
1   8  28  56  70  56  28   8   1

This reveals a very specific pattern. The outermost term (always 1) shares factors with all numbers within its row. The next term (the terms counting 1,2,3,4,5,6...) shares factors will all terms enclosed between its two pairs (but not necessarily the same factor*!). The next inner terms seem to share this property and it seems to carry on throughout the entire system. Another principle following is that any adjacent digit within a power of 11 (when ignoring carry over) shares some factor (other than 1) except in cases where one of the digits = 1. Granted, we remove the two end-digits, so the case of 1 is ignored.

note*in the case of odd powers, the inner end-digits are the common factor.

Now, is there some principle that states that if every adjacent digit shares factors then the number is a non-prime number? I really don't know! I can conclude absolutely that all prime number* powers of 11 truncated and shifted in this manner are divisible. In fact, the factor is n where n is the power 11 is raised to! The non-prime powers** are indeterminate.

note* numbers that are powers of prime numbers will also work such as 4, or 8 or 9.

note** I cannot entirely validate that they dont exist, merely that I cannot derive a proof for their existence. These will involve multiplication of multi-term binomials. As such, I cannot verify them easily. However, prime based n's will have coefficient divisibility.

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protected by user26857 Nov 18 at 11:00

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