Coefficients of a series expansion under modular arithmetic

Question

I am reading a paper by Ramanujan on partitions and need help understanding an intermediate step he uses in a proof. He writes:

... all the coefficients in $(1-x)^{-5}$ are multiples of 5, except those of $1$, $x^5$, $x^{10}$, $...$, which are congruent to 1: that is to say $$ \dfrac{1}{(1-x)^5} \equiv \dfrac{1}{1-x^5} \pmod 5 $$ or$$ \dfrac{1-x^5}{(1-x)^5} \equiv 1 \pmod 5 $$

The coefficients can easily be verified to have this property, we have the series expansion:

$$ (1-x)^{-5} = 1 + 5x + 15x^2 + 35x^3 +70x^4 + 126x^5 + \dots $$

Here's how I would approach it. For the series expansion I can write:

$$ \begin{align} (1-x)^{-5} &= \sum_{k=0}^\infty \binom{k+4}{4}x^k\\ &= \sum_{k=0}^\infty \dfrac{(k+4)!}{4!k!} x^k\\ &= \sum_{k=0}^\infty \dfrac{(k+4)(k+3)(k+2)(k+1)}{4!}x^k\\ &= \sum_{k=0}^\infty \dfrac{24 + 50 k + 35 k^2 + 10 k^3 + k^4}{24}x^k \end{align} $$

and looking at the coefficients modulo 5:

$$ \dfrac{24 + 50 k + 35 k^2 + 10 k^3 + k^4}{24} \equiv 1-k^4 \pmod 5 $$

and I can see that $k \equiv 0$ gives the coefficient congruent to 1 while any other value of k results in the coefficient being congruent to 0, the desired result. Still, I'm confused at the relevance/meaning of the congruences that Ramanujan wrote. Can someone help me understand what he meant? I feel like I'm missing something.

Answer 1

I hope you know the following

Theorem: If $p$ is a prime number then $p$ divides the binomial coefficients $\binom{p}{a}$ for $a=1,2,\dots, (p-1)$.

Therefore we have $$(1-x)^{5}\equiv 1-x^5\pmod{5}$$ and taking reciprocals $$\frac{1}{(1-x)^{5}}\equiv \frac{1}{1-x^5}\pmod {5}$$

The key here is the congruence of formal power series with integer coefficients. Let $\sum_{n=0}^{\infty} a_nx^n, \sum_{n=0}^{\infty} b_nx^n$ be two formal power series with integer coefficients. Then we say that $$\sum_{n=0}^{\infty} a_nx^n\equiv \sum_{n=0}^{\infty} b_nx^n\pmod{m} $$ if $a_n\equiv b_n\pmod {m} $ for all $n$.

Such congruences follow usual rules like one can add / subtract same formal series to each side of a congruence or multiply both sides of a congruence by same formal series and the resulting congruence remains valid.

Similarly we can take reciprocals of both sides of a congruence if resulting formal series have integer coefficients (prove it!). Note that this requires the formal series involved have their constant terms equal to $1$ so that taking reciprocals leads to integer coefficients.

---

Ramanujan was a master of such manipulation with formal series and products and he used this approach extensively. My guess here is that you encountered this technique in proving partition congruence $$p(5n+4)\equiv 0\pmod{5}$$