## Task 28/82

Determine all prime pairs $(p ; q)$ for which $\binom{p}{q}$ is also a prime number!
Due to $\binom{p}{q}=1$ for $p=q,\binom{p}{q}=0$ for $p<q$. For $p>q$, $\binom{p}{q}$ contains the prime factor $p$ in the numerator, but not in the denominator.

Therefore, $\binom{p}{q}$ is divisible by $p$ without a remainder. The condition of the problem is thus satisfied if and only if $\binom{p}{q}=p$. Now,

$$
\binom{p}{q}=\frac{p(p-1)(p-2) \ldots(p-q+1)}{1 \cdot 2 \cdot \ldots \cdot q}=p
$$

is true if and only if

$$
(p-1)(p-2) \ldots(p-q+1)=q!
$$

From the uniqueness of the prime factorization, it follows that the non-1 factors on the right and left sides of the equation are pairwise equal:

$$
p-q+1=2, \ldots, p-1=q
$$

Thus, $p=q+1$. This equation is (in prime numbers $p ; q$) only solvable for $q=2, p=3$. Indeed,

$$
\binom{p}{q}=\binom{3}{2}=\binom{3}{1}=3
$$

No further solutions can exist based on the solution path.