Task 5. Find all increasing arithmetic progressions with prime number terms that have the property: the number of terms in the progression is finite and greater than the difference of the progression.
Solution: We will prove that the sought progressions are 2,3 and $3,5,7$.

Consider a progression with a difference of 1 that satisfies the condition of the problem. It has at least two terms. One of them must be even, and the only even prime number is 2. There are no prime numbers less than 2, so 2 is the first term of the progression. The second term is 3, and the third term (4) is not a prime number.

Consider a progression with a difference of 2 that satisfies the condition. It contains at least 3 terms; let the first three be $k-2, k, k+2$. Clearly, regardless of the remainder $k$ gives when divided by 3, one of these terms is divisible by 3. The prime number divisible by 3 is 3. It is easy to see that 3 can only be the first term of the progression; we get $3,5,7$. The fourth term is 9, a composite number.

Consider the progression $a_{1}, a_{2}, \ldots, a_{n}$ with a difference $d \geq 3$. The numbers $d-1$ and $d$ are coprime, so all prime divisors of $d-1$ do not divide $d$. Let $p$ be one of these prime divisors. Consider the part of the progression $a_{2}, a_{3}, \ldots, a_{p+1}$. (This is a part of the progression since $p<d<n$.) All remainders modulo $p$ of the numbers $a_{2}, a_{3}, \ldots, a_{p+1}$ are different. (Indeed, if

$$
\left.\begin{array}{l}
a_{i}=a_{1}+(i-1) d=s p+r \\
a_{j}=a_{1}+(j-1) d=t p+r
\end{array}\right\} \Longrightarrow(t-s) p=(j-i) d
$$

Since $j-i<p$, the right side of the last equation is not divisible by $p$, while the left side is. Contradiction.) Therefore, one of the considered remainders is 0. This means that one of the numbers $a_{2}, a_{3}, \ldots, a_{p+1}$ is divisible by $p$. However, this is impossible, since each of these numbers is prime and greater than $p$ (since it has a number at least 2 and is therefore greater than $d$).