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I got a question to show that :

If $p$ is prime number, then $$x^p - x \equiv x(x-1)(x-2)(x-3)\cdots (x -(p-1))\,\,\text{(mod }\,p\text{)}$$

Now I got 2 steps to show that the two polynomials are congruent:

  1. Show they have same roots.
  2. Show the leading term's coefficients on both sides are the same.

For step 2, I know it is obvious $x^p$ has coefficient 1 on both sides. For step 1, but I am not sure how to find all the solutions for $x^p -x$ modulo $p$.

Bill Dubuque
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morgan
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2 Answers2

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Fermat's little theorem says that $x^p\equiv x$ modulo $p$. So the left hand side of your equation is $0$. The right hand side of your equation is also $0$, because one of the numbers

$$x,x-1,x-2,\dots,x-(p-1)$$

is divisible by $p$.

J.R.
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It is not clear if we are asked to prove equality as polynomial functions (i.e. they have the same values) or as formal polynomials (i.e. they have the same coefficients).

For the former, by little Fermat both sides are $\equiv 0\,$ for $\,x\equiv 0,1,\ldots, p\!-\!1.\,$ And, for the latter, let $\,f(x) =$ LHS $-$ RHS. Since LHS and RHS have same lead coefs we deduce that $f$ has degree $< p,$ and, by little Fermat, $\,f\,$ has $\,p\,$ distinct roots $\,1,2,\ldots p\,$ in the field $\,\Bbb Z/p.\,$ Therefore $\,f\,$ must be the zero polynomial (i.e. all coefs $\equiv 0)$ since a nonzero polynomial over a field (or domain) has no more roots than its degree (a proof follows by iteratively applying the Factor Theorem, and using the fact that the difference of any two distinct roots is nonzero so cancellable). See also the BiFactor Theorem.

Remark $ $ This may fail in non-domains, e.g. mod $\,8\!:\,\ x^2-1\,$ has $\,4\,$ roots $\, x\equiv \pm1,\,\pm 3,\,$ i.e. ${\rm odd}^2\equiv 1\pmod 8$

Bill Dubuque
  • 272,048
  • WOW, it is quite a formal proof! Am I reading a paper in math? Thank you so much. – morgan Feb 24 '14 at 20:15
  • @user131041 Was it your idea to use Steps 1,2 or was that suggested to you? If the latter, then you are probably being asked to show they are equal as formal polynomials, i.e. that both sides have congruent coefficients. – Bill Dubuque Feb 24 '14 at 20:20
  • yeah, you got it right. I am suggested to first find all the roots of the LHF, and then compare the leading coefficients on the both sides. – morgan Feb 24 '14 at 20:30
  • Interestingly enough, this leads to an easy proof of Wilson's Theorem: Since $x^{p-1}-1$ has $p-1$ zeroes $1,2,3,\cdots,p-1$, and $\pmod p$ is a domain, $x^{p-1}-1\equiv (x-1)(x-2)(x-3)\cdots(x-(p-1))$. Set $x=0$. – chubakueno Feb 26 '14 at 04:34
  • @chubakueno Yes, that is one of the standard proofs. – Bill Dubuque Feb 26 '14 at 04:35