1. What is the greatest length of an arithmetic progression of natural numbers

$a_{1}, a_{2}, \ldots, a_{n}$ with a difference of 2, in which for all $k=1,2 \ldots, n$ all numbers $a_{k}^{2}+1$ are prime
Solution.

All members of the progression must be even numbers. Consider the remainders when divided by 5. Since the common difference of the arithmetic progression is 2, the remainders in the progression will appear in the order

$\ldots 2,4,1,3,0,2,4,1,3,0, \ldots$ This means that among the terms of the progression, there will be numbers of the form $a_{k}=5 t+2$ and $a_{k}=5 t+3\left(t-\right.$ natural). For these numbers, we calculate $a_{k}^{2}+1=$ $25 t^{2}+20 t+5$ and $a_{k}^{2}+1=25 t^{2}+30 t+10$. We have found that the specified terms of the progression $a_{k}=5 t+2$ and $a_{k}=5 t+3$ yield expressions $a_{k}^{2}+1$ that are divisible by 5 and are not prime. The longest set of terms in the arithmetic progression that does not contain numbers of the form $a_{k}=5 t+2$ and $a_{k}=5 t+3$ consists of two numbers. Therefore, the number of consecutive terms of the progression that meet our criteria cannot exceed two. The only case where a sequence of three numbers can be obtained is by including the number 5 in $a_{k}^{2}+1$. Indeed, the numbers $2,4,6$ form an arithmetic progression and yield the corresponding sequence of prime numbers 5, 17, 37. Answer. $n=3$.