# Proof of Babbage's Theorem?

Does anyone know how to prove that

p*a choose p*b = a choose b (mod p^2)

Additional information:

http://en.wikipedia.org/wiki/Wolstenholme%27s_theo...

(Proof of the theorem is on there but too difficult to understand)

### 4 Answers

- gianlinoLv 79 years agoFavorite Answer
Note that "ap choose bp" is the coefficient of x^(bp) in the binomial expansion of (1+x)^(ap). Write this as ( (1+x)^p )^a.

Expanding (1+x)^p we get 1 + p Q(x) + x^p, where Q is a polynomial with integer coefficients, since p is prime.

We'll say that 2 polynomials are equal, if all their coefficients are equal mod p^2.

Now (1+x)^(pa) = (1 + x^p + pQ(x))^a

= (1+x^p)^a + a (1+x^p)^(a-1)*pQ(x) + terms in p^2....

Since pQ(x) = (1+x)^p - 1 - x^p, we have

(1+x)^(pa) = (1+x^p)^a + a (1+x^p)^(a-1) * ((1+x)^p - 1 - x^p)

= (1+x^p)^a * ( 1 - a ) + a (1+x^p)^(a-1) * (1+x)^p

We are interested in the coefficient of x^(bp). In (1+x)^p only the terms 1 and x^p will come in since all other powers are multiples of p.

So the coefficient of x^(bp) in

(1+x^p)^a * ( 1 - a ) + a (1+x^p)^(a-1) * (1+x)^p

is the same as its coefficient in (1+x^p)^a * ( 1 - a + a) = (1+x^p)^a,

which is ..... a choose b. That's it.

Edit @ Ben: No we don't need; what counts here are the exponents.

Since the exponents in (1+x^p)^a and (1+x^p)^(a-1) are all multiples of p, and in the end we want the coefficient of x^(bp) only the terms of (1+x)^p, whose exponents are multiples of p will count, namely 1 and x^p. Right?

- BenLv 69 years ago
The proof given in the wikipedia article uses some basic group action ideas; if you're not familiar with them, then I could try to cook up an algebraic proof. But let me just elaborate on the proof given in wiki:

Let G be the ath power of the cyclic group C_p. Let A be a set with p*a elements. Partition A into sets A_1,A_2,...,A_a, each with p elements. Think of G as acting on A, with each component C_p acting independently on one of the subsets cyclically.

Now consider a different space X: the set of all subsets of A with size p*b. Our group (C_p)^a has a natural action on this space induced by the previously described action: start with a B in X, and let phi in (C_p)^a send B to the set of all phi(b) for b in B. That is, act on each element of B individually, then collect the images together into a new set. It should be clear that the result still has p*b elements. Now we want to find the size of the orbits of this new action (the wiki article states the size without justification; it's not hard, but perhaps a bit confusing). So let B be in X. How many sets does it get sent to under various phi? Consider the intersection of B with each A_i. If the intersection is empty or all of A_i, then different phi don't look different in that component. If the intersection is nonempty, then there are p different images in that component (since the restriction of the action to each A_i is regular). So let k denote the number of i such that B intersect A_i is nonempty and not all of A_i (the article calls these incomplete rings), and we have that the size of the orbit of B is p^k. (We do this for each B.)

Okay, we're finally ready to get what we want. Basic counting techniques tell us that the size of X is C(pa, pb). Group action theory tells us that this is the sum of the sizes of the orbits of any action on X. So we have that C(pa, pb)=sum(p^k), where the sum is taken over all orbits of the action described above, and k is the number defined above for some (any) B in the orbit. Modulo p^2, the only surviving terms in the sum come from those with k=0 and those with k=1.

If the orbit of B has k=0, then each intersection of B with an A_i is either empty or all of A_i. There are of course C(a, b) of these (just choose b of the sets A_i to be full).

The orbit of B cannot have k=1, since |B| is divisible by p.

Hence the sum has C(a, b) terms which are 1, and the rest are zero modulo p^2, and we have Babbage's theorem.

Let me know if there's anything I can elaborate on here to help.

EDIT: Ah, gianlino seems to have an algebraic proof.

[gianlino, I can't figure out why toward the end you say that we only care about the 1 and the x^p from the term (1+x)^p...don't we need to at least concern ourselves also with the px and px^(p-1) terms?]

EDIT2: Oh, I see. I misread the phrase "all other exponents are multiples of p" as meaning any x^blah was p*x^blah. You're correct.

- MargaretLv 44 years ago
@hogzeye: What he means is that if God is omnipotent (all-powerful) then he can will 1 to equal 0. What I can't understand is where Homer Simpson comes into this. Anyway, I think you've answered your own question, asker. It is not up to humans to say one can equal zero. God has absolute authority over the a priori world so only he can can decide.