Given $3$ coprime polynomials $f,g,h\in \mathbb{C}[x]$ such that $f+g=h$. Prove that the number of different zeroes of the polynomial $fgh$ is bigger then the $\deg f$. Have no idea about this one.
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I could be totally off, and I'm not terribly experienced in this field, but by the fundamental theorem of algebra, doesn't a polynomial $f$ of degree $n$ have $n$-many zeros in $\mathbb{C}$? If that's true, then $deg(fgh)>deg(f)$ and the answer follows from there. – malxmusician212 Dec 12 '16 at 18:18
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1The zeroes don't have to be different (in the fundamental theorem). – Dec 12 '16 at 18:18
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@menag ah, thanks. Figured I couldn't be right since I didn't use $f+g=h$ – malxmusician212 Dec 12 '16 at 18:20
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This is known as Mason-Stothers theorem – Wojowu Jan 31 '17 at 19:18
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Here's a official solution translated from Serbian
Since $f+g=h$,from $(f,g,h)=1$ we have that $f,g,h$ are coprime in pairs such that the number of different zeroes of $fgh$ is equal to $\deg f-\deg \gcd(f,f')+\deg g-\deg \gcd(g,g')+\deg h-\deg \gcd(h,h')$
(polynomial from $\mathbb{C}[x]$ of degree $n$ has $n$ complex roots; if $k\in\Bbb{N}$ is a multiplicity of a root $\alpha$ of $f$ then $\alpha$ is a root of $f'$ with multiplicity $k-1$)
Let $F=fg'-f'g$ (if $\alpha$ is a root of $f$ with multiplicity $k\in \Bbb{N}$,from (f,g)=1 it's not a root of $g$ so it's a zero of $F$ with multiplicity $k-1$ so $F\not= 0$). Then $F=f(h-f)'-f'(h-f)=fh'-hf'$ and $F=(h-g)g'-(h-g)'g=hg'-h'g$,and because $(f,g,h)=1$,it follows that $\gcd(f,f')\gcd(g,g')\gcd(h,h')\mid F$ it follows $\deg F<\deg g+\deg h$ $$\deg fgh= \deg f+(\deg g+\deg h)-(\gcd(f,f')\gcd(g,g')\gcd(h,h'))\geq \deg f+(\deg g+\deg h)-\deg F>f$$
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