Given $N$ is the product of $m$ distinct primes, how many $K(K+1)$ are divisible by $N$?

Question

Let $N$ be the product of two distinct primes $P_1$ and $P_2$.

Let $S$ be a set of all $K<N$, such that $N|K(K+1)$.

For example, if $N$ is $5\cdot7=35$ then $S=\{14,20\}$.

The size of $S$ in this case is $2$, and this is true for any distinct primes $P_1$ and $P_2$ that we choose.

---

Now let's make it a little more general:

Let $N$ be the product of $m$ distinct primes $P_1,\dots,P_m$.

Let $S$ be a set of all $K<N$, such that $N|K(K+1)$.

What is the size of $S$?

Answer 1

Let $r_i$ be the remainder of $K$ modulo $P_i$, Then $r_1\in\{0,-1\}$. Altogether there are $2^m$ allowed sequences $(r_1,\ldots,r_m)$. By the Chinese Remainder Theorem, there is exactly one $K$ among $1,\ldots,N$ with those remainders. Since you disallow $K=N$, the answer is $2^m-1$.

(Your example is inaccurate, for $N=5\cdot7$ there are 3 possible values: 14,20 and 34.)