Irreducibility of a simple polynomialShow $x^6 + 1.5x^5 + 3x - 4.5$ is irreducible in $mathbb Q[x]$.Determine...
voltage of sounds of mp3files
Is there a measurement for the vocal speed of a song?
Go Pregnant or Go Home
Teaching indefinite integrals that require special-casing
MaTeX, font size, and PlotLegends
Is the destination of a commercial flight important for the pilot?
How do I define a right arrow with bar in LaTeX?
At which point does a character regain all their Hit Dice?
Opposite of a diet
Personal Teleportation as a Weapon
Why are on-board computers allowed to change controls without notifying the pilots?
Bash method for viewing beginning and end of file
quarter to five p.m
How was Earth single-handedly capable of creating 3 of the 4 gods of chaos?
Select empty space and change color in vector
I'm in charge of equipment buying but no one's ever happy with what I choose. How to fix this?
Time travel short story where a man arrives in the late 19th century in a time machine and then sends the machine back into the past
Is there an Impartial Brexit Deal comparison site?
Can I convert a rim brake wheel to a disc brake wheel?
Increase performance creating Mandelbrot set in python
Why did Kant, Hegel, and Adorno leave some words and phrases in the Greek alphabet?
Lay out the Carpet
Is there any reason not to eat food that's been dropped on the surface of the moon?
Why is delta-v is the most useful quantity for planning space travel?
Irreducibility of a simple polynomial
Show $x^6 + 1.5x^5 + 3x - 4.5$ is irreducible in $mathbb Q[x]$.Determine whether the polynomial $x^2-12$ in $mathbb Z[x]$ satisfies an Eisenstein criterion for irreducibility over $mathbb Q$Proving Irreducibility of $x^4-16x^3+20x^2+12$ in $mathbb Q[x]$Constructibility of roots of a polynomialEisenstein's criterion for polynomials in Z mod pProving irreducibility; What is this method and what is the logic behind it?Irreducibility of special cyclotomic polynomial.Irreducibility of a Polynomial after a substitutionIrreducibility of Non-monic Quartic Polynomials in Q[x]Irreducible monic polynomial in $mathbb{Q}[x]$
$begingroup$
For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.
What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.
For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.
A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
asked 1 hour ago
JonHalesJonHales
520311
$endgroup$
add a comment |
$begingroup$
For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.
What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.
For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.
A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
asked 1 hour ago
JonHalesJonHales
520311
$endgroup$
2
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
1 hour ago
add a comment |
$begingroup$
For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.
What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.
For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.
A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
asked 1 hour ago
JonHalesJonHales
520311
$endgroup$
For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $mathbb{Q}$.
What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.
For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.
A thought I had was to try and consider when $mathbb{Q}(sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.
abstract-algebra field-theory irreducible-polynomials
abstract-algebra field-theory irreducible-polynomials
asked 1 hour ago
JonHalesJonHales
520311
asked 1 hour ago
JonHalesJonHales
520311
asked 1 hour ago
JonHalesJonHales
520311
asked 1 hour ago
JonHalesJonHales
520311
asked 1 hour ago
JonHalesJonHales
520311
520311
2
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
1 hour ago
add a comment |
2
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
1 hour ago
2
2
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
1 hour ago
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Notice that we are trying to reduce that polynomial by this way:
$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$
We need:
$$2a-b^2=0$$
$$b=sqrt{2a}$$
But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:
$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$
Which is also known as Sophie Germain Identity.
answered 1 hour ago
EurekaEureka
22611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
1 hour ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
1 hour ago
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
1 hour ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
1 hour ago
add a comment |
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.
answered 1 hour ago
egregegreg
185k1486206
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3163711%2firreducibility-of-a-simple-polynomial%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that we are trying to reduce that polynomial by this way:
$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$
We need:
$$2a-b^2=0$$
$$b=sqrt{2a}$$
But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:
$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$
Which is also known as Sophie Germain Identity.
answered 1 hour ago
EurekaEureka
22611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
1 hour ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
1 hour ago
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
1 hour ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
1 hour ago
add a comment |
$begingroup$
Notice that we are trying to reduce that polynomial by this way:
$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$
We need:
$$2a-b^2=0$$
$$b=sqrt{2a}$$
But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:
$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$
Which is also known as Sophie Germain Identity.
answered 1 hour ago
EurekaEureka
22611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
1 hour ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
1 hour ago
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
1 hour ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
1 hour ago
add a comment |
$begingroup$
Notice that we are trying to reduce that polynomial by this way:
$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$
We need:
$$2a-b^2=0$$
$$b=sqrt{2a}$$
But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:
$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$
Which is also known as Sophie Germain Identity.
answered 1 hour ago
EurekaEureka
22611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Notice that we are trying to reduce that polynomial by this way:
$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$
We need:
$$2a-b^2=0$$
$$b=sqrt{2a}$$
But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:
$$x^4+4k^4=(x^2-2kx+2k^2)(x^2+2kx+2k^2)$$
Which is also known as Sophie Germain Identity.
answered 1 hour ago
EurekaEureka
22611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
EurekaEureka
22611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
EurekaEureka
22611
answered 1 hour ago
EurekaEureka
22611
22611
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
1 hour ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
1 hour ago
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
1 hour ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
1 hour ago
add a comment |
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
1 hour ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
1 hour ago
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
1 hour ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
1 hour ago
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
1 hour ago
$begingroup$
I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$
$endgroup$
– Sil
1 hour ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
1 hour ago
$begingroup$
@Sil It's pretty easy with complex factorization. And it can also be done with a system of equations.
$endgroup$
– Eureka
1 hour ago
1
1
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
1 hour ago
$begingroup$
@Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$.
$endgroup$
– Ethan MacBrough
1 hour ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
1 hour ago
$begingroup$
@EthanMacBrough I understand, my point though was that things like that should be in answer itself.
$endgroup$
– Sil
1 hour ago
add a comment |
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.
answered 1 hour ago
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.
answered 1 hour ago
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.
answered 1 hour ago
egregegreg
185k1486206
$endgroup$
Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt{2a}x)^2=
(x^2-sqrt{2a}x+a)(x^2+sqrt{2a}x+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbb{R}$. By uniqueness of factorization, this is a factorization in $mathbb{Q}[x]$ if and only if $2a$ is a square in $mathbb{Q}$.
answered 1 hour ago
egregegreg
185k1486206
answered 1 hour ago
egregegreg
185k1486206
answered 1 hour ago
egregegreg
185k1486206
answered 1 hour ago
egregegreg
185k1486206
185k1486206
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3163711%2firreducibility-of-a-simple-polynomial%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible.
$endgroup$
– Sil
1 hour ago