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Given a prime $p$ and some $k,t\in\Bbb{Z}^{+}$, when does the congruence equation $${x \choose k} \equiv t\pmod {p}$$ have an integer solution?

Is there some necessary and sufficient condition about $p,k,t$?

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  • $\begingroup$ @JCAA I changed my question. Thank you. $\endgroup$ – MathEric Jun 30 at 17:58
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By Lucas's theorem, $${x \choose k} \equiv \prod_j {x_j \choose k_j} \mod p$$ where the base-$p$ representations of $x$ and $k$ have digits $x_j$ and $k_j$ respectively, (with extra $0$'s as needed in $k_j$ when $k$ has fewer digits than $x$). If all $k_j = 0$ or $p-1$, for example, ${x_j \choose k_j} \equiv 0$ or $1 \mod p$, and then${x \choose k} \equiv 0$ or $1 \mod p$.

So for a given $k$ and $p$, with $[k_d, \ldots, k_0]$ the base $p$ representation of $k$, the possible nonzero values $t$ are the products $t = \prod_{j=0}^d t_j \mod p$ where $t_j \in \{{0 \choose k_j} \mod p, \ldots, {p-1 \choose k_j} \mod p\}$.

For example, with $p=5$ and $k = 123 = 443_5$, ${x \choose 4} \equiv 0$ or $1 \mod 5$, and ${x \choose 3} \equiv 0, 1$ or $4 \mod 5$, so the possibilities for ${x \choose 123}$ are $0, 1$ and $4 \mod 5$.

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