Moonmathpi496 wrote: What is the largest possible length of an arithmetic progression of positive integers $ a_{1}, a_{2},\cdots , a_{n}$ with difference $ 2$, such that $ {a_{k}}^{2}+1$ is prime for $ k = 1, 2, . . . , n$? Take any three consecutive terms $ a,a+2,a+4$ and look modulo 5 : If $ a=0\pmod{5}$, then $ (a+2)^{2}+1=0\pmod{5}$ and the second is divisible by $ 5$. If $ a=1\pmod{5}$, then $ (a+2)^{2}+1=0\pmod{5}$ and the second is divisible by $ 5$. If $ a=2\pmod{5}$, then $ a^{2}+1=0\pmod{5}$ and the first is divisible by $ 5$. If $ a=3\pmod{5}$, then $ a^{2}+1=0\pmod{5}$ and the first is divisible by $ 5$. If $ a=4\pmod{5}$, then $ (a+4)^{2}+1=0\pmod{5}$ and the third is divisible by $ 5$. So, among any three consecutive terms, one of the $ {a_{k}}^{2}+1$ is divisible by $ 5$, so is not prime (except if it is equal to 5). So the maximum is two, except if the number $ 2$ is in the sequence. And in this case we can have a three-elements sequence : $ 2,4,6$ which gives the three primes $ 5,17,37$. The answer is $ 3$.

Answer:$ n = 3$. Solution: Denote $ p_{i}= a_{i}^{2}+1$.If $ a_{i}\equiv 2,-2(mod 5)$ then $ 5|a_{i}^{2}+1$.If $ a_{i}> 2$ and $ a_{i}\equiv 2,-2(mod 5)$ then $ p_{i}$ isn't prime number.So it is obviously that among any three numbers $ a,a+2,a+4$ exist one number that $ \equiv 2,-2(mod 5)$.Now if in the progression isn't exist $ 2$,then $ n\leq 2$.If $ a_{1}= 2$ then $ a_{2}= 4$,$ a_{3}= 6$.And $ p_{1}= 5,p_{2}= 17,p_{3}= 37$. Edit:I was too slow.I'm sorry pco.

Anyone have the whole problem set(with the distribution for each grades) and solution for russia 2002 & 2003? Thanks