# Congruence equation with binomial coefficient

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 $$p,k,t$$?

• @JCAA I changed my question. Thank you. – MathEric Jun 30 at 17:58

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 $$4 \mod 5$$.