Let us consider integers $a,b$ and $c$ such that all of them are greater than 1. I am trying to figure out whether the following three divisibility conditions can be satisfied together
- $a|b^{a-1}$
- $ab|c$
- ${{ab}\choose{a}}|b^{a-1}c$
or, in other words, if we can find such integers. I started with the 3rd condition and plugged in a value for $a$, solved for $c$ and plugged in values for $b$ but $c$ turned out to be non-integer for the values I tried.
Correct me if I have misunderstood the question, but it seems quite straightforward when you observe the fact that $c$ never came in the divisor side. First, we can set $a=b=x$ so that the first condition holds. Next, we need $x^2 \mid c$ and $\binom{x^2}{x} \mid x^{x-1}c$. Both these conditions will trivially hold if $c=x^2 \cdot \binom{x^2}{x}$. Thus, such solutions exist and one example is: $$(a,b,c)=\bigg(x,x,x^2 \cdot \binom{x^2}{x}\bigg)$$ It might be helpful to sometimes start from basic statements with few of the variables absent, rather than directly jumping to the crux of the arguments (3rd statement).