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I'm working on a problem where I have a quadratic polynomial $p$ with integer coefficients and a situation where I have three numbers $x_0,x_1,x_2$ such that $f(x_0)=f(x_1)=f(x_2)=1$. Now there is a remark on the book that

A quadratic polynomial cannot assume the same value for three different values of $x$. This is really an application of the fundamental theorem of algerba.

How is this result true? The conclusion is that if $f(x)=ax^2+bx+c$ and $f(x_0)=ax_0^2+bx_0+c=1, f(x_1)=ax_1^2+bx_1+c=1$ and $f(x_2)=ax_2^2+bx_2+c=1$, then $f$ must be identically $1$, but I cannot figure out why.

Walker
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  • Are you familiar with the theorem that, for a polynomial $p(x)$, if $p(a)=0$, then $(x-a)$ divides $p(x)$? – Sambo Sep 30 '22 at 15:35
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    Function $f(x)-1$ would have three zeroes, which is impossible. – Intelligenti pauca Sep 30 '22 at 15:36
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    That remark from the book is misleading. That result isn't really an application of the fundamental theorem of algebra. While you can involve the fundamental theorem of algebra in the argumentation of that result, it is not necessary. Moreover, the fundamental theorem of algebra requires some "continuity properties" of the real numbers (like real polynomials of odd degree having at least one root), while the result being proven only needs algebraic properties, in particular that $ab=0$ implies that either $a=0$ or $b=0$. – plop Sep 30 '22 at 15:40
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    You can prove the conclusion simply by solving the system of three linear equations that you have, in which the unknowns are $a,b,c$. You will get that $a=b=0$ and $c=1$. The condition that $x_0,x_1,x_2$ are different appears since to get this solution you will need to divide by $(x_0-x_1)(x_1-x_2)(x_2-x_3)$ to get $a,b,c$. – plop Sep 30 '22 at 15:44
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    I upvoted this question, because this is deeper than it gets credit for. I will post an answer later to elaborate. For now however, that a nonconstant polynomial of degree $n$ has only $n$ roots, implicitly relies on the fact that the set $\mathbb{C}$ of complex numbers is a field. – Mike Sep 30 '22 at 16:40

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You have three equations. If you subtract the first two, you get

$$f(x_0) - f(x_1) = a(x_0^2-x_1^2) + b(x_0-x_1) = 0.$$

Since $x_0 \neq x_1$, you can divide both sides by $x_0-x_1$ to get

$$a(x_0+x_1) + b = 0$$

so that

$$x_0 + x_1 = -\frac{b}{a}.$$

Similarly you get

$$x_0 + x_2 = -\frac{b}{a}.$$

If you subtract these last two equations, you get $x_1-x_2= 0$, so they're equal. Which is a contradiction.

B. Goddard
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