How prove this $f(x)=g(x)=h(x)=0$?

Question

let polynomial $f(x),g(x),h(x)\in C[x]$, and such $$f^2(x)=xg^2(x)+xh^2(x),$$

prove or disprove $$f(x)=g(x)=h(x)=0$$

I know solve this follow problem let polynomial $f(x),g(x),h(x)\in R[x]$, and such $$f^2(x)=xg^2(x)+xh^2(x),$$

prove or disprove $$f(x)=g(x)=h(x)=0$$

then I guess when $f,g,h\in C[x]$ is true? Thank you

What if $g(x) = 1$, $h(x)=i$ and $f(x)=0$? – Siméon Nov 26 '13 at 10:02 — Siméon, Nov 26 '13 at 10:02

score 2 · Accepted Answer · answered Nov 26 '13 at 10:36

2

Actually, this won't be true at all in $\Bbb C$. It might even be the case that each one of $f,g,h$ is non zero. There follows an example: $$ f(x) = 2e^{i\pi/4}x,\qquad g(x) = ix+1, \qquad h(x) = x+i. $$

answered Nov 26 '13 at 10:36

Siméon

10,664
1
21
54

egreg · Answer 2 · 2013-11-26T10:22:08.530

0

We can, without loss of generality, assume one of $g$ and $h$ is nonzero.

Let $g(x)=x^a g_1(x)$, $h(x)=x^a h_1(x)$ in such a way that either $x$ doesn't divide $g_1$ or doesn't divide $h_1$.

Then $x$ doesn't divide $g_1(x)^2+h_1(x)^2$ (unless …). But $$ f(x)^2=x^{2a+1}(g_1(x)^2+h_1(x)^2) $$ so …

edited Nov 26 '13 at 10:22

answered Nov 26 '13 at 10:04

egreg

238,574

You made a mistake with "then $x$ doesn't divide $g_1(x)^2+h_1(x)^2$". See my comment on the OP. – Siméon Nov 26 '13 at 10:08
@Siméon I was silently assuming that $g$ and $h$ are non zero. – egreg Nov 26 '13 at 10:20
Still, take $g(x) = 1$ and $h(x) = i$ for instance. – Siméon Nov 26 '13 at 10:21
@Siméon What's the problem in that case? We'd simply have $a=0$. – egreg Nov 26 '13 at 10:23
The problem is that $x$ does divide $g_1(x)^2+h_1(x)^2$. – Siméon Nov 26 '13 at 10:24
@Siméon OK, I'll edit. – egreg Nov 26 '13 at 10:25

How prove this $f(x)=g(x)=h(x)=0$?

2 Answers2