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From Barbeau's Polynomials:

  • (a) Is it possible to find a polynomial, apart from the constant $0$ itself, which is identically equal to $0$ (i.e. a polynomial $P(t)$ with some nonzero coefficient such that $P(c)=0$ for each number $c$)?

And then I thought about 2 hypothesis:

  1. P(c-c)
  2. I thought about a polynomial such as $ax^2+bx+c=0$, then I could make a polynomial with $a=1$, $b=-x$, $c=0$ which would render $x^2+(-x)x=0$. I tested it on Mathematica with values from $-10$ to $10$ and it gave me $0$ for all these values.

When I went for the answer, I've found:

               enter image description here

When I went to the answer, I couldn't understand it, can you help me? I'm trying to know what's he doing in this answer, I guess it's a way to prove it, but It's still intractible to me. You can explain me or recommend me some thing for reading. I'll be happy if you also tell me something about my hypothesis. Thanks.

  • 1
    Your idea in (2) doesn't work because the coefficients of a polynomial have to be independent of the variable. We choose $a,b,c$ and *then* form a polynomial $ax^2 + bx + c$; you can't choose $b=-x$ because at the time you choose $a,b,c$ there is no such thing as $x$ yet.2012-09-19
  • 1
    Your idea in (1) doesn't work either, because after substituting every instance of $c-c$ with $0$, you get the zero polynomial.2012-09-19
  • 0
    Incidentally, for an even better understanding of this concept and proof you might want to have a look at the polynomial $x^{p^n}-x$ over the (finite) field $GF(p^n)$ - for instance, the polynomial $x^2-x$ over the two-element field $GF(2)$ - and try to understand how things can break down there.2012-09-25

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