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In this answer: https://math.stackexchange.com/a/652909

Ted equates mod $p$ the coefficients of $$\sum_{n=0}^{pa} \binom{pa}{n} x^n$$ and $$\sum_{i=0}^{a} \binom{a}{i} x^{pi}$$ to get that $$\binom{pa}{pb} \equiv \binom{a}{b} \mod p$$

But one of the summations has $pa$ terms, and the other summation has $a$ terms, so what does it mean to equate the coefficients mod $p$? Shouldn't it be that several different elements of $\sum \binom{pa}{n} x^n$ will collapse into one element of $\sum \binom{a}{i} x^{pi}$ since $x^{pb} \equiv x^{p^kb} \mod p$ so that after modding by $p$, the coefficients of each $x^{p^kb}$ will contribute to the coefficient of $x^{pb}$ i.e. $$\binom{pa}{b} + \binom{pa}{pb} + \binom{pa}{p^2b} \dots \equiv \binom{a}{b} \mod p$$

Is Ted's proof complete? And if not, what would a complete proof look like?

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yeons
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1 Answers1

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In the original post, this was not my proof, but clark's. I merely edited some typos in it. But I can address your concerns.

First, you cannot write something like $x^{pb} \equiv x^{p^k b} \mod{p}$. Not sure how you got that, so I'll just say this: When working with polynomials mod $p$, the exponents do not reduce modulo $p$. The exponents are ordinary (nonnegative) integers. You can only equate powers that are exactly equal as integers (you cannot reduce them modulo $p$). Only the coefficients can be reduced modulo $p$.

So what do we get when we equate coefficients of equal powers of $x$? On the right side (the second sum in your question above), the term $i=b$ gives $\binom{a}{b}$ as the coefficient of $x^{pb}$. This corresponds to the term $n=pb$ in the first sum, which has coefficient $\binom{pa}{pb}$. These can be equated modulo $p$, giving $\binom{pa}{pb} \equiv \binom{a}{b} \mod p$.

To address your concern that the first sum has $pa$ terms and the second sum only $a$ terms, what's happening is that most of the terms in the first sum are 0. If $n$ is not divisible by $p$, then $\binom{pa}{n}$ is divisible by $p$ and therefore becomes 0 when working modulo $p$. The argument in the previous paragraph is a proof of this (since the second sum obviously has no terms that are not multiples of $p$), but you can also prove it directly.

Ted
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