I'm just reading a paper which, on its page 3, Application 8, claims the following:
$$\binom{k+sp}{j}\equiv\binom{k}{j}\pmod{p}$$
where $p\ge 1$, $s\ge 1$, $k\ge 1$ and $p\not\mid j$ (actually, it claims that this holds even in a generalized setting where binomial coefficients are subscripted by some $q$ for which $q=1$ retrieves the ordinary binomial coefficients).
However, when I let $k=5$, $j=4$, $s=1$, $p=3$, then
$$\binom{k+sp}{j}=\binom{8}{4}=70\not\equiv 5=\binom{5}{4}=\binom{k}{j}\pmod{p}.$$
Can anyone confirm that I'm right or am I completely missing something? (And if I am in fact correct, the other congruence indicated in Application 8 is probably false, too?)
Some assumptions are missing, which your example violates. You need to assume $0\leq s,k,j<p$. I'm not seeing anything in the paper in question which would enforce this (they do seem intent on $k,j$ simply being coprime). So there's either a convention that enforces this that's been overlooked/unstated, or there is an error in the paper.
As it turns out, the $q=1$ case for your example boils down to a special case of Lucas's Theorem. That special case being that for any prime $p$ and non-negative integers $n,n_0,r,r_0$ that are all less than $p$ we have
$${{np+n_0}\choose{rp+r_0}}\equiv {n\choose r}{n_0\choose r_0}\mod p.$$
For your example, with $p=3$ you should instead be looking at $$70={8\choose 4}={{2p+2}\choose{p+1}}\equiv {2\choose 1}{2\choose 1}=4\equiv 1\mod p.$$ Which is true.
Expressions of the form $np+n_0$ for such $n,n_0$ are simply the base $p$ expansion of a number between $0$ and $p^2-1$. The fully general form of Lucas's theorem applies to arbitrary base $p$ expansions. If you apply proper base $p$ expansions to the results of your paper things might work out, but I've not yet attempted to consider if this fixes the $q\neq 1$ case.