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Let $f(x)=x^4+6x^2+1 \in \mathbb{Z}[X]$. We have to prove that $f$ is reducible modulo $p$ for all prime $p$.

I am not sure how to proceed. Please give me some hints so that I can try solving the problem.

Edit I:

I have tried with the following primes.

  1. For p=2; $f(x)$ reduces to $x^4+1 = (x^2+1)^2$ since $\bar{6}=\bar{0}=\bar{2}$. Hence $f(x)$ is reducible mod 2.

  2. For p=3; $f(x)$ also reduces to $x^4+1=(x^2+x+2)(x^2+2x+2)$ in $\mathbb{Z}_3[X]$

Edit II: Following JyrkiLahtonen's comment I am adding a few more things:

A. Taking $y=x^2;\ f(x)$ becomes $y^2-6y+1$. Solving the equation $y^2-6y+1=0$ we get $$ y=-3 \pm 2 \sqrt{2} $$ Now replacing $y$ by $x^2$ we get the following $$ x=\pm\sqrt{-3+2\sqrt{2}}=\pm(i-i\sqrt{2}) \ \ \ \text{and} \ \ \ x=\pm\sqrt{-3-2\sqrt{2}}=\pm(i+i\sqrt{2}) $$ Hence we have obtained all the roots of the polynomial $f(x)$; say $x_1,x_2,x_3,x_4$.

B. The three ways of expanding this quartic are as follows:

  1. $\{(x-x_1)(x-x_2)\}\{(x-x_3)(x-x_4)\}=(x^2+1-2ix)(x^2+1+2ix)\\ = (x^2+1)^2-(2ix)^2 \\ = (x^2+1)^2-2^2(-1)x^2$

  2. $\{(x-x_1)(x-x_4)\}\{(x-x_2)(x-x_3)\}=(x^2-1-2\sqrt{2}ix)(x^2-1+2\sqrt{2}ix)\\ = (x^2-1)^2-(2\sqrt{2}ix)^2\\ =(x^2-1)^2-2^2(-2)x^2$

  3. $\{(x-x_1)(x-x_3)\}\{(x-x_2)(x-x_4)\}=(x^2+3+2\sqrt{2})(x^2+3-2\sqrt{2})\\ =(x^2+3)^2-(2\sqrt{2})^2\\ =(x^2+3)^2-2^2(2)$

C. Now the final piece.

$f(x)$ can be factored in any of the above 3 ways. Consider any prime $p$.

  1. If -1 is square element in $\mathbb{F}_p$; there exist $a \in \mathbb{F}_p$ such that $-1=a^2$. So factoring $f(x)$ as in form B.1. Hence considering $f(x)$ in $\mathbb{F}_p[X]$ we have $$ f(x)=(x^2+1)^2-2^2(-1)x^2=(x^2+1)^2-(2ax)^2=(x^2+1-2ax)(x^2+1+2ax) $$ Hence $f(x)$ is reducible modulo $p$ if $-1$ is a square element.

  2. If 2 is square element in $\mathbb{F}_p$, where $p$ is an odd prime; there exist $a \in \mathbb{F}_p$ such that $2=a^2$. So we factor $f(x)$ as in form B.3. Now considering $f(x)$ in $\mathbb{F}_p[X]$ we have $$ f(x) =(x^2+3)^2-2^2(2)=(x^2+3)^2-(2a)^2=(x^2+3+2a)(x^2+3-2a) $$ Hence we get $f(x)$ is reducible modulo $p$ if $2$ is a square element.

  3. Now if both $-1$ and $2$ are non square elements in $\mathbb{F}_p$ then $-2$ is a square element in $\mathbb{F}_p$ (follows by using Legendre symbol as mentioned in the comment by Jyrki Lahtonen.) Hence factoring $f(x)$ as in the form B.2 we get that $f(x)$ is reducible modulo $p$.

Hence we have $f(x)$ is reducible modulo $p$ for all $p$.

This is a special case of a more general infinite class of polynomials with similar property. I have written the general case here.

Saikat
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    What primes have you done so far? Edit your question to show us your methods and progress. – Ted Shifrin Jun 30 '21 at 04:30
  • There are examples in the right margin under Related. Hint: $\sqrt{-1}+\sqrt{-2}$. – Jyrki Lahtonen Jun 30 '21 at 04:43
  • Thanks for the edit. Unfortunately it is still difficult to give a good answer. One way forward would be to use field extensions and minimial polynomials. What do you know about such themes? Another would be to split this according to which of $-1$, $-2$, $2$ is a quadratic residue modulo $p$ (my first comment points in that direction). Like in Hagen von Eitzen's answer here. – Jyrki Lahtonen Jun 30 '21 at 06:32
  • You need to tell us about the theoretical tools you have at your disposal. If I just bluntly state that $\Bbb{F}_p(i,\sqrt2)$ is a degree $\le2$ splitting field of this polynomial, will it solve the question for you? We need background information, or context. – Jyrki Lahtonen Jun 30 '21 at 06:33
  • @JyrkiLahtonen It would be very nice if you could tell me something using basic tools. Although If there are no such basic tools available. Then it will be ok if you give me an answer using the field extensions (if you tell it elaborately then it will be very helpful, as I have not learned field extensions well.) – Saikat Jun 30 '21 at 07:39
  • Are you familiar with quadratic residues? The plan is then to A) Observe that the complex zeros of this polynomial are $\pm i\pm i\sqrt2$. B) Write down and expand the three ways of factoring this quartic as a product of two quadratics. C) Observe that, after expanding, one of those ways has polynomials with coefficients involving integers and $\sqrt{-1}$, another with coefficients involving integers and $\sqrt{-2}$, and the last with coefficients involving integers and $\sqrt{2}$. D) If you know that at least one of $-1,-2,2$ is a quadratic residue modulo $p$, you can use that. – Jyrki Lahtonen Jun 30 '21 at 07:47
  • Not unlike here where the resulting factors have coefficients involving one of $\sqrt2$, $\sqrt3$, $\sqrt6$. – Jyrki Lahtonen Jun 30 '21 at 07:52
  • The type of factorization that shows modulo $p$ really depends on which of $-1,-2,2$ have a square root modulo $p$. The bit of theory you need is that at least one of those has a modular square root modulo any prime $p$. I don't know of away of proving this without using either quadratic residues or basic facts about extensions of finite fields. – Jyrki Lahtonen Jun 30 '21 at 07:55
  • @JyrkiLahtonen I have computed A), B) and C) and added to my question. Please see, if possible. – Saikat Jun 30 '21 at 08:43
  • @JyrkiLahtonen I read Hagen Von Eitzen's answer. Your comment on D) has a similarity with his answer. But I am unable to understand how can I apply this technique here. – Saikat Jun 30 '21 at 08:49
  • You made a sign error. You should get $y=-3\pm2\sqrt2$. That is two negative numbers with imaginary square roots. Or may be you intended $y=-x^2$? The calculations and factorizations are probably ok. Didn't check. – Jyrki Lahtonen Jun 30 '21 at 08:50
  • Have you covered quadratic residues? That is the key to understanding Hagen von Eitzen's solution. – Jyrki Lahtonen Jun 30 '21 at 08:50
  • I have read the definition of quadratic residues. Is that enough? I am trying to read his answer once again. – Saikat Jun 30 '21 at 08:53
  • Yes sorry I made a typing mistake. we get $y=-3\pm2\sqrt{2}$ – Saikat Jun 30 '21 at 08:57
  • @JyrkiLahtonen I have understood Hagen Von Eitzen's answer and also have found how to apply it here. The only problem I am facing in his answer is the following: He writes if -1, 2 are not squares in $\mathbb{F}_p$ then -2 is a square. How do we obtain this result? – Saikat Jun 30 '21 at 09:06
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    If you know the Legndre symbol, then all you need is the rule $$\left(\frac{-1}p\right)\left(\frac2p\right)=\left(\frac{-2}p\right).$$ If both $-1$ and $2$ are quadratic non-residues then this implies that $-2$ is a quadratic residue. So one of those three always is a quadratic residue. If this does not make sense to you, the next step could be to use the existence of a primitive root modulo $p$. – Jyrki Lahtonen Jun 30 '21 at 09:10
  • You could post most of it as an answer, because it evolved into one :-) – Jyrki Lahtonen Jun 30 '21 at 11:47
  • @Saikat The given answer looks correct to me, although I will go over it once. – Sarvesh Ravichandran Iyer Jul 20 '21 at 06:40
  • @JyrkiLahtonen I have found an infinite class of polynomials with this property, of which this is a special case. If possible please have a look at it. I have added the link in the bottom of the post. – Saikat Jul 20 '21 at 08:22

1 Answers1

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Here is a bit of a different approach.

For $p=2$, we can factor $$ \begin{align} x^4+6x^2+1 &\equiv(x-1)^4&\pmod2 \end{align} $$ For the odd primes, where $p\in\{1,3,5,7\}\pmod8$, we will use the results from this answer, to show that $-1$ is a quadratic residue mod $p$ iff $p\equiv1\pmod4$, and this answer, to show that $2$ is a quadratic residue mod $p$ iff $p\equiv\pm1\pmod8$. Together, those answers show that $-2$ is a quadratic residue mod $p$ iff $p\equiv1,3\pmod8$.

Since $p\equiv1,5\pmod8$ implies there is a $q$ so that $q^2\equiv-1\pmod p$, we can factor $$ \begin{align} x^4+6x^2+1 &=\left(x^2+1\right)^2-(-4)x^2\\ &\equiv\left(x^2-2qx+1\right)(x^2+2qx+1)&\pmod p\\ \end{align} $$ Since $p\equiv1,3\pmod8$ implies there is a $q$ so that $q^2\equiv-2\pmod p$, we can factor $$ \begin{align} x^4+6x^2+1 &=\left(x^2-1\right)^2-(-8)x^2\\ &\equiv\left(x^2-2qx-1\right)(x^2+2qx-1)&\pmod p\\ \end{align} $$ Since $p\equiv1,7\pmod8$ implies there is a $q$ so that $q^2\equiv2\pmod p$, we can factor $$ \begin{align} x^4+6x^2+1 &=\left(x^2+3\right)^2-8\\ &\equiv\left(x^2+3-2q\right)(x^2+3+2q)&\pmod p\\ \end{align} $$

robjohn
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