# A Proof of Fermat's Little Theorem

1. Dec 6, 2009

### 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

2. Dec 6, 2009

### 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.

3. Dec 6, 2009

### 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 begining n=1.

Last edited: Dec 7, 2009
4. Dec 12, 2009

### 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? $$(K+1)^p - (k+1) \equiv 0 \bmod p.$$

5. Dec 12, 2009

### nomather1471

Last edited by a moderator: May 4, 2017
6. Dec 13, 2009

### 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.

Last edited: Dec 13, 2009
7. Dec 13, 2009

### nomather1471

Is there any problem at my proof?

8. Dec 13, 2009

### 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!