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Bob tells me that the Identity Theorem is the three following statements

  1. If a polynomial has infinitely many roots, then it is equal to $0$.

  2. If two polynomials satisfy $P(x)=Q(x)$ for infinitely many $x$, then the two polynomials are equal.

  3. If two polynomials of at most degree $n$ satisfy $P(x)=Q(x)$ for $n+1$ values of $x$, then the two polynomials are equal.

I do not know how to prove these; I think the Factor, Remainder, and Division Theorems will be useful.

I tried using Fundamental Theorem of Algebra, but it did not get me anywhere.

Blue
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user627514
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    Who is Bob and why his statements should be of interest? – user Dec 27 '18 at 13:10
  • Use number $3$ to prove the second one... Then use the second one to prove the first. Then all that remains is proving the third one. Alternatively, prove these in the reverse order – abiessu Dec 27 '18 at 13:13
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    what have you tried with the Fundamental Theorem of Algebra? What does it tell you about a non-constant polynomial? – Jonas Lenz Dec 27 '18 at 13:14
  • @user Probably, "Bob" refers to http://www.mathdoctorbob.org/ – lisyarus Dec 27 '18 at 13:16
  • Hint : A polynomial of degree $n$ can have at most $n$ distinct roots – Peter Dec 27 '18 at 13:22
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    Note that it matters that the polynomials have coefficients over a field (or a domain). So real or rational or integer coefficients will do. But without this condition we have modulo $8$ the polynomial $x^2-1$ has four roots $1,3,5,7$ and factorisations $(x-1)(x-7)\equiv (x-3)(x-5) \equiv x^2-1$, so factorisation is not unique. – Mark Bennet Dec 27 '18 at 13:24
  • @Peter Is $P (x)\equiv0$ a polynomial? – user Dec 27 '18 at 13:26
  • @user Yes, called the "zero-polynomial" – Peter Dec 27 '18 at 13:28
  • @peter Then probably your statement is valid only for non-zero polynomials. – user Dec 27 '18 at 13:31
  • @user I assumed that the degree $n$ is $0$ or positive and not $-\infty$, which is usually considered to be the degree of the zero-polynomial. – Peter Dec 27 '18 at 13:32
  • You need to specify the coefficient ring else it may fail, e.g. $,x^2 = 1,$ has $4$ roots over $,\Bbb Z/8 = $ integers mod $8$ since $,{\rm odd}^2\equiv 1\pmod{8}\ $ (and such rings do frequently occur in number theory - your tag). – Bill Dubuque Dec 27 '18 at 15:24

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For polynomials with coefficients in $\mathbb C$, the three facts are easy consequences of the fundamental theorem of algebra, which says that a nonzero polynomial has $n$ ( not necessarily distinct) roots, where $n$ is the degree.

For example, $n+1$ would be too many roots, leaving the zero polynomial as the only way out.

Bill Dubuque
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  • The field doesn't really matter. Over any field, a non-zero polynomial of degree $n$ has no more than $n$ roots, since each root corresponds to a degree $1$ factor of the polynomial. – mr_e_man Mar 30 '23 at 21:58