Find all the positive integers a, b, and c such for which $\binom{a}{b} \binom{b}{c} = 2\binom{a}{c}$ I tried an the following equivalence from a different users post from a while back
that stated the following.
$\binom{a}{b} \binom{b}{c} = \binom{a}{c} \binom{a-c}{b-c}$
where does this equivalence come from?
After applying this equivalence and trying to hammer it out algebraically 
I end up with...
$\frac{(a-c)!}{(b-c)!(a-b)!}=2$
Not any closer than when I didn't use the equivalence.
How can I solve this?

You correctly arrived at recognizing that $\binom{a-c}{b-c}=2$
Now, the only remaining step is to recognize that the only binomial coefficient which equals two is $\binom{2}{1}$, thus $a-c=2$ and $b-c=1$, so they form a triple $(c+2,c+1,c)$
Why is this?  You can prove it by strong induction noting that $\binom{n}{0}=1, \binom{n}{n}=1$ and $\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$ which for $n\geq 3$ is by inductive hypothesis either $0+1$ or is the sum of two positive numbers, one of which is greater than or equal to $2$, and is thus greater than or equal to three.
Alternatively, you can note that $\binom{n}{1}=\binom{n}{n-1}=n$ and binomial coefficients are monotonic up until the midpoint., that is $n=\binom{n}{1}<\binom{n}{2}<\binom{n}{3}<\dots<\binom{n}{\lfloor n/2\rfloor}$, similarly $\binom{n}{\lceil n/2\rceil}>\dots>\binom{n}{n-2}>\binom{n}{n-1}=n$
