# factor-square property (FSP) of polynomials

The Factor Square Property (FSP) is the divisibility of the polynomial $f(x^2)$ by $f(x)$.

1. Is $x^2+x+1$ the only FSP irreducible polynomial of degree $2$ ?

2. Are there other linear polynomial besides $x$ and $x-1$ with FSP?

3. Do we have other FSP irreducible polynomials of degree $3$ or $4$? Any of these have integer coefficients??

4. Are there any other observations you can make about polynomials with FSP?

• What is the factor-square property? – Chris Eagle Mar 26 '13 at 20:25
• f(x) has FSP if f(x) is a factor of f(x2) example F(x) = x-1 has FSP because x-1 is a factor of x2-1. similarly, h(x) =x has FSP because x is a factor of x2. – sam Mar 26 '13 at 20:50
• Please, TeX. It exists for a reason. – Lord_Farin Mar 26 '13 at 22:18
• @sam : The definition of FSP, that $f(x)$ divides $f(x^2)$, does not appear to be a standard definition. In my opinion the definition of this term FSP (or factor square property) should appear in the main body of the question, not just as a response to someone (Chris Eagle) having asked what it means. – coffeemath Mar 27 '13 at 7:51
• As pointed out by someone in another question, this is from math.osu.edu/ross/app/RossProb13.pdf – Erick Wong Mar 31 '13 at 4:35

A google search of "factor square property" led only to this present question. So recall from comment that a polynomial $f(x)$ which divides $f(x^2)$ is said to be FSP or to have the factor square property.

This addresses [1], and shows an irreducible FSP quadratic must be $ax^2+ax+a$, which depending on what "irreducible" means might also imply $a=1$. We also address [3], giving a source of examples.

Let the quadratic be $p=ax^2+bx+c$, so its value at $x^2$ is $q=ax^4+bx^2+c$. If $p$ is to be a divisor of $q$ let the other factor be $dx^2+ex+f.$ Equating coefficients gives equations

[1] $ad=a,$

[2] $ae+bd=0,$

[3] $af+be+cd=b,$

[4] $bf+ce=0,$

[5] $cf=c.$

Now we know $a,c$ are nonzero (else $p$ is not quadratic, or is reducible). So from [1] and [5] we have $d=f=1.$ Then from [2] and [4] we obtain $ae=ce.$ Here $e=0$ leads to $b=0$ from either [2] or [4], and [3] then reads $a+c=0$, so that $p=a(x^2-1)$ which is reducible. So we may assume $e$ is nonzero, and also $a=c.$

At this point, [2] and [4] say the same thing, namely $ae+b=0.$ So we may replace $b=-ae$ in [3] (with its $c$ replaced by $a$) obtaining $a+(-ae)e+a=-ae,$ which on factoring gives $a(2-e)(e+1)=0.$ The possibility $e=2$ then leads after some algebra to $2a+b=0$ and $p=a(x-1)^2$ which is reducible, while the possibility $e=-1$ leads to $a=b$ and then $p=ax^2+ax+a$ as claimed.

For some higher degree examples, for any odd prime $p$ the cyclotomic polynomial $f(x)=(x^p-1)/(x-1)$ is seen to be FSP because $$\frac{f(x^2)}{f(x)}=\frac{x^{2p}-1}{x^2-1}\cdot \frac{x-1}{x^p-1} \\ =\frac{x^p+1}{x+1}.$$ For $p=5$ this is the polynomial $$x^4+x^3+x^2+x+1.$$ Note that for odd prime $p$ these cyclotomic polynomials are known to be irreducible (there is a simple Eisenstein proof using a change of variables).

• what about the polynomials with the real number coefficient and the complex number coefficient. Here is the link of my question: math.stackexchange.com/questions/2128306/… – Mclalalala Feb 5 '17 at 6:30
• Thank you if you can help me do some further exploration – Mclalalala Feb 5 '17 at 6:30
• @Mclalalala The argument above about finding the quadratic irreducibles with FSP is independent of whether the coefficients are real or complex. I do not claim to have found all possible higher degree polynomials with FSP, only a collection of them. – coffeemath Feb 5 '17 at 17:18