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A Proof of Fermat's Little Theorem

  1. Dec 6, 2009 #1
    Hi guys,

    I've been reading a chapter from a book and I've been attempting a question which uses the binomial theorem to prove Fermat's Little Theorem. The question goes as follows:

    Let p be a prime number:
    i) Show that if r, s are positive integers such that r divides s, p divides r and p does not divide s, then p divides [tex]\frac{r}{s}[/tex].
    ii) Deduce that p divides the binomial coefficient [itex]\displaystyle \binom{p}{k}[/itex] for any k such that [tex]1 \leq k \leq p-1[/tex].
    iii) Now use the binomial theorem to prove by induction on n that p divides n[tex]^{p}[/tex] - n for all positive integers n. Hence deduce Fermat's Little Theorem.

    I can handle the first two parts of the question, but I think I may not have showed them in a way which leads onto being able to prove the third part. For i) I said that since r divides s I can express s as some multiple of r, which gets me r in both the numerator and the denominator of the fraction and thus since p divides r p must divide the fraction. I did a similar approach for the second part except using the expanded [tex]\frac{p!}{k!(p-k)!}[/tex] form.

    Could someone please help me finish off the rest of the question? I'm really stumped...

  2. jcsd
  3. Dec 6, 2009 #2
    I imagine that the binomial theorem will come in during the induction step, when you have to say something about (n+1)^p - (n+1).

    You may want to revise the statement of the question at I), because something doesn't look right. Divisibility is transitive, so if p|r and r|s then p|s, it cannot be that "p does not divide s" as stated. Maybe it is s who divides r and not the other way around, since the question uses the fraction [tex]\frac r s[/tex] and apparently expects it to be an integer.
  4. Dec 6, 2009 #3
    r divides s, p divides r and p does not divide s,

    This is what looks wrong to Dodo, just ignore it.

    Now use the binomial theorem to prove by induction on n that p divides n^p - n.

    What is said about is the Key Statement and a giveaway. For p a prime, just start with the usual begining n=1.
    Last edited: Dec 7, 2009
  5. Dec 12, 2009 #4
    If nobody else wants to say anything, I feel the need to elaborate.

    What is the basis? [tex] 1^p-1 \equiv 0 \bmod p. [/tex]
    What is the induction hypothesis?[tex] K^p-K \equiv 0 \bmod p.[/tex]

    What to prove? [tex] (K+1)^p - (k+1) \equiv 0 \bmod p. [/tex]
  6. Dec 12, 2009 #5
    http://www.hizliupload.com//viewer.php?id=475631.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  7. Dec 13, 2009 #6
    Sure! Correct! If you take an example (x+y)^3 = X^3 + 3x^2(y) + 3x(y^2) + y^3. The fact remains that [tex]\frac {p!}{0!p!} [/tex] will divide out p, and similarly for the last term. Every other term contains p only in the numerator.

    Thus from the problem, [tex] (K+1)^P -(K+1)\equiv K^p+1^p -(K+1) \equiv K^p-K \equiv 0 \bmod p [/tex]
    The last step by the induction hypothesis.
    Last edited: Dec 13, 2009
  8. Dec 13, 2009 #7
    Is there any problem at my proof?
  9. Dec 13, 2009 #8
    nomather1471: Is there any problem at my proof?

    No! I just thought it was a little lengthy writing out all those coefficients. Plus it was a litle difficult to bring it up.

    I guess I should have left it alone!
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