Show that there does not exist a polynomial $f \in \mathbb{R}[x]$ such that $f^2(x) = f(x) \cdot f(x) = 1 + x + x^3$.

Question

I really have no idea where to begin and would appreciate all help I can get to solve this.

i tried to use the information from this link Is there a polynomial $f\in \mathbb Q[x]$ such that $f(x)^2=g(x)^2(x^2+1)$ - because it resembles my doubt. But I couldn’t do it.

for one thing, $f(x)^2$ should be non-negative for all $x\in\Bbb R$, and $1+x+x^3$ is not — J. W. Tanner, Dec 28 '20 at 06:32
What polynomial, when squared, results in a polynomial of odd degree? — paulinho, Dec 28 '20 at 06:33
That link is about $\mathbb Q[x]$, so considerations are different from those of $\mathbb R[x]$ — J. W. Tanner, Dec 28 '20 at 06:53
You recieved 4 answers to your question. Is any of them what you needed? If so, consider accepting the best answer and upvoting all useful answers you got. That's how the site works. — 5xum, Dec 29 '20 at 07:47

score 2 · Answer 1 · answered Dec 28 '20 at 06:49

2

Alternative approach:

$f(x)^2$ should be non-negative for all $x\in\mathbb R$.

What about $1+x+x^3$, particularly when $x$ is a negative number with large magnitude?

answered Dec 28 '20 at 06:49

J. W. Tanner

score 1 · Answer 2 · answered Dec 28 '20 at 06:33

1

Hint:

If the degree of $f$ is $n$, what is the degree of $f^2$?

answered Dec 28 '20 at 06:33

5xum

if $f=x^n \Rightarrow f^2=x^{2n}$. but I can’t show the steps. – Fernando Sousa Dec 28 '20 at 06:40
@FernandoSousa I didn't ask "if $f=x^n$". I asked "If the degree of $f$ is $n$". Also, the proof that the degree of a product of two polynomials is the sum of their degrees can be found in any textbook on introductory calculus. – 5xum Dec 28 '20 at 06:48

score 1 · Answer 3 · answered Dec 28 '20 at 06:54

1

Any

$f(x) \in F[x], \tag 1$

where $F$ is any field, may be written

$f(x) = \displaystyle \sum_0^n f_i x^i = f_nx^n + \sum_0^{n - 1}f_ix^i, \; f_i \in F; \tag 2$

then

$f(x) \cdot f(x) = f_n^2x^{2n} + \text{lesser degree terms}; \tag 3$

it follows that $f(x) \cdot f(x)$ is always of even degree; but

$\deg \; x^3 + x + 1 = 3, \; \text{odd}, \tag 4$

so . . .

Note that

$F = \Bbb R \tag 5$

needn't hold to attain the above result.

answered Dec 28 '20 at 06:54

Robert Lewis

1

Since $f(x)=1+x+x^3$ when $f(x)=0$ has no real roots, then $f^2(x)= 1+x+x^3$, so $f \notin \mathbb{R}$, since there is no $f(-1) \cdot f(-1)=f(-1)^2=-1=\sqrt{-1}=i$ and $i \in \mathbb{C}$. Can I think of it this way? – Fernando Sousa Dec 28 '20 at 07:47

3 Answers3