1

Here's my attempt, I'm almost there but I'm stuck:

Using a hint, I wrote the modular reduction:

Reducing the coefficients modulo $2$ gives: $\left [ f \right ]_2=x^4+x^2+x=x(x^3+x+1)$.

Reducing the coefficients modulo $3$ gives: $\left [ f \right ]_3=x^4+x^3+x-1$

If $f$ were reducible then either it decomposes as $(x-x_0)g(x)$ (in which case $x_0$ is a root of $f$, therefore $\left [ x_0 \right ]_3$ is a root of $\left [ f \right ]_3$ but it is easy to verify $\left [ f \right ]_3$ has not roots so this is impossible) or it decomposes as $p(x)q(x)$, in which case $\left [ f \right ]_2=\left [ p \right ]_2\left [ q \right ]_2$. But I don't know how to find a contradiction here.

Thanks in advance.

John11
  • 1,569
  • Modulo $3$ the polynomial is $(x^2+x+2)(x^2+1)$, hence reducible. This gives no contradiction. – Dietrich Burde May 13 '19 at 18:14
  • 2
    As $x^3+x+1$ is irreducible modulo $2$, then the polynomial can only factor as a linear times a quadratic. For it to have a linear factor some number dividing $4$ must be a root.... – Angina Seng May 13 '19 at 18:16
  • 1
    Please double check that the coefficients are all correct. – Bill Dubuque May 13 '19 at 18:23
  • @BillDubuque I just did and they are indeed correct. Is it not supposed to be irreducible in $\mathbb{Z}[t]$? – John11 May 13 '19 at 18:27
  • 1
    Ok, then probably you are meant to use ideas like LStU suggested since the first small prime that works is $p = 17$, and (shifted) Eisenstein also does not work. Were you taught any other irreducibility tests? – Bill Dubuque May 13 '19 at 18:31
  • @BillDubuque no, just Eisenstein and reducibility. – John11 May 13 '19 at 20:03

3 Answers3

1

The polynomial is irreducible modulo $17$. This is (a little bit) easier than to check over the integers. First, there is no root modulo $17$, and then writing the polynomial as $(x^2+ax+b)(x^2+cx+d)$ quickly gives a contradiction modulo $17$.

Substituting $x$ by $x-1$ we obtain $f=x^4-3x^2+9x-11$, which is a bit easier to handle. First we have $c=-a$ and then $bd=-11=6$. This means, we have $(b, d)=(1,6),(2,3), (3,2), (4,10), (5,8), (6,1), (7,13), (8,5),\ldots $, which is enough to solve the last two equations for $a$.

Dietrich Burde
  • 130,978
  • This is likely quite difficult, unless you know some specific optimization. Are you sure it is feasible? Did you try to complete it or are you just guessing? – Bill Dubuque May 13 '19 at 18:33
  • Yes, I tried it now. It treats some cases, which is not so elegant, but elementary. – Dietrich Burde May 13 '19 at 19:18
  • But how is one supposed to know to try the prime $p = 17$? That's a lot of work in itself excluding all the smaller primes. It's much easier to use info gained from the degrees of factors mod small primes – Bill Dubuque May 13 '19 at 19:41
0

Hint: Write out what it means for the product of two $2$nd degree polynomials to have product equal to $f$. Get an system of equations in the coefficients. Check whether the system is consistent.

0

HINT.-You have $$f(1)=11\\f(-1)=-11\\f(3)=233\\f(5)=1231\\f(-9)=3821\\f(-17)=64613\\f(25)=455171$$ Then there is seven integers $n=1,-1,3,5,-9,-17,25$ such that $f(n)$ is prime. This gives $3$ integers more than the degree of the polynomial.

Note that just one prime $f(n)$ gives a strong enough probability that $f(x)$ be irreducible.

If I remember correctly a result of mine established that if there are two integers more than the degree of the polynomial giving $ f (n) $ prime then $ f (x) $ is irreducible. It could be that instead of $ 2 $ it has been $ 3 $ but that is exhibited here, with $ 3 $ more integers. If you want, take this just as a comment.

Piquito
  • 29,594
  • These ideas go back to Kronecker, e.g. see here. – Bill Dubuque May 13 '19 at 20:57
  • Two or three years after I held this in Congresses, it appeared, with a slight improvement, as an anonymous note in a prestigious portal. If you want I look for the dates and the portal I do not remenber now (I believe that in one of Weinstein). – Piquito May 13 '19 at 23:18
  • If you are interested in the history of these very old ideas you can learn about that from the citations in the links I gave. The general philosophy led to (inefficient) polynomial factorization algorithms by Bernoulli, Schubert, Kronecker, Hausmann, et al. – Bill Dubuque May 13 '19 at 23:25
  • Who said that the algorithm used here is efficient? There are irreducible polynomials that from $1$ to $1000$ or more give no primes. Can you build one? There is a person who yesterday put a downvote without justifying it to an answer on elliptical curves (my PhD subject in France) that I gave 27/11/2017, what do you think is the intention of this person? – Piquito May 14 '19 at 16:55
  • My remark about efficiency concerns only said ancient polynomial factorization algorithms (as compared to modern algorithms). As for your other answer, I have no idea since I don't know who did that. – Bill Dubuque May 14 '19 at 17:02
  • One needs "chance" to use this algorithm. In general is not so useful as one could desire. Note however that an old conjecture, still not proven, is that all irreducible gives $f(n)$ as infinitely many primes (what is a generalization of the beautiful theorem of the arithmetic progression). Regards. – Piquito May 14 '19 at 19:48
  • See here for more on that. – Bill Dubuque May 14 '19 at 19:55
  • Interesting. A lady proved the conjecture, for some particular quadratic cases in the AMS Bulletin some years ago. She used the prime number theorem for this. I do not remember the name of the lady or the number of the Bulletin. – Piquito May 14 '19 at 23:39
  • @Bill Dubuque: Betty Garrison “Polynomials with large numbers of prime” American Mathematical Monthly Volume 97, Number 4, April, 1990 – Piquito May 15 '19 at 11:33