A Proof of Fermat's Little Theorem

[2^Oscar]

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 \frac{r}{s}.
ii) Deduce that p divides the binomial coefficient \displaystyle \binom{p}{k} for any k such that 1 \leq k \leq p-1.
iii) Now use the binomial theorem to prove by induction on n that p divides n^{p} - 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 \frac{p!}{k!(p-k)!} form.

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

 

[dodo]

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 \frac r s and apparently expects it to be an integer.

 

[robert Ihnot]

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 beginning n=1.

 

[robert Ihnot]

If nobody else wants to say anything, I feel the need to elaborate.

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

What to prove? (K+1)^p - (k+1) \equiv 0 \bmod p.

 

[nomather1471]

http://www.hizliupload.com//viewer.php?id=475631.jpg

 

[robert Ihnot]

Sure! Correct! If you take an example (x+y)^3 = X^3 + 3x^2(y) + 3x(y^2) + y^3. The fact remains that \frac {p!}{0!p!} will divide out p, and similarly for the last term. Every other term contains p only in the numerator.

Thus from the problem, (K+1)^P -(K+1)\equiv K^p+1^p -(K+1) \equiv K^p-K \equiv 0 \bmod p
The last step by the induction hypothesis.

 

[nomather1471]

Sure! Correct! If you take an example (x+y)^3 = X^3 + 3x^2(y) + 3x(y^2) + y^3. The fact remains that \frac {p!}{0!p!} will divide out p, and similarly for the last term. Every other term contains p only in the numerator.

Thus from the problem, (K+1)^P -(K+1)\equiv K^p+1^p -(K+1) \equiv K^p-K \equiv 0 \bmod p
The last step by the induction hypothesis.

Is there any problem at my proof?

 

[robert Ihnot]

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!