corollary of the partition congruence

Question

I was going though the paper of Ramanujan entitled Some properties of p(n), the number of partitions of n (Proceedings of the Cambridge Philosophical Society, XIX, 1919, 207 – 210). He states he found the proofs of p(5m + 4) ≡ 0 (mod 5) and p(7m + 5) ≡ 0 (mod 7). Then he says p(35m + 19) ≡ 0 (mod 35) can be deduced as corollary. I did not get how do derive the third congruence p(35m + 19) ≡ 0 (mod 35) from p(5m + 4) ≡ 0 (mod 5) and p(7m + 5) ≡ 0 (mod 7), where p(n) unrestricted partition function.

Answer 1

We are given that

$\dots$ he found the proofs of $p(5m + 4) \equiv 0 \pmod 5$ and $p(7m + 5) \equiv 0 \pmod 7$.

The two statements can be written as

$$ n \equiv 4\pmod 5 \implies p(n)\equiv 0\pmod 5 $$

and

$$ n \equiv 5\pmod 7 \implies p(n)\equiv 0\pmod 7. $$

The Chinese remainder theorem applied to $n$ implies that the two conditions on $n$ combined are equivalent to $ n \equiv 19\pmod{35}$. Similarly, the two conditions on $p(n)$ combined are equivalent to $p(n)\equiv 0\pmod{35}$. The conclusion is

$$ n \equiv 19\pmod{35} \implies p(n)\equiv 0\pmod{35}.$$