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$p_{1}= 2 \equiv 2 (mod 5)$ $p_{2}= p_{1}+1 = 3 \equiv 3 (mod 5)$ Note that $(p_{n}-1)p_{n}+1 = p_{n+1}$ If $p_{n}\equiv 2 (mod 5)$ and $p_{n+1}\equiv 3 (mod 5)$, then $p_{n+2}\equiv (p_{n+1}-1)p_{n+1}+1 \equiv (3-1)*3+1 \equiv 2 (mod 5)$, and $p_{n+3}= (p_{n+2}-1)p_{n+2}+1 \equiv (2-1)*2+1 \equiv 3 (mod 5)$. So if any two consecutive terms are $2$ and $3$ in $mod 5$, respectively, the next two must also be. So in $mod 5$, the sequence goes $2, 3, 2, 3, 2, 3, 2, 3$ ... and no term is divisible by $5$.

neelnal wrote: Starting...

no because you have $p_{1}p_{2}...p_{n-1}+1$ and now $p_{1}p_{2}...p_{n-1}$ @ZutonForce: the problem is, you take the largest prime divisor of $p_{1}p_{2}...p_{n-1}+1$ and not $p_{1}p_{2}...p_{n-1}+1$ itself. So the mod will be different. I will post the sol-n in a few days if nobody solves the problem.

rem wrote: neelnal wrote: Starting...

no because you have $p_{1}p_{2}...p_{n-1}+1$ and now $p_{1}p_{2}...p_{n-1}$ I was not trying to say that your problem was wrong. I was just eliminating the possibility the p1p2...p(n-1) + 1 has a 0 as its last digit since p1=2, making p1p2...p(n-1) + 1 odd.

Oh sorry I did not quite understand what you wrote. Yes that's a start: the last digit cannot be 0.

Some more thoughts:

Please do not post ideas that lead nowhere and are irrelevant to the problem. The problem does have a very nice and simple solution. And by the way, the problem is number theory and it is not proposed, it appeared on Iberoamerican olympiad 1987, question 5.

but it's very easy and nice! 2,3 are the first two terms in the sequence and a prime can appear only once in $p_{1},p_{2},...$ so for 5 to appear we need $p_{1}p_{2}\cdots p_{n}=5^{n}-1$ at some stage but 4 doesn't divide LHS

Yes that's the nice solution I was talking about. Good job! I really like this beautiful rpoblem.

Albanian Eagle wrote: but it's very easy and nice! 2,3 are the first two terms in the sequence and a prime can appear only once in $p_{1},p_{2},...$ so for 5 to appear we need $p_{1}p_{2}\cdots p_{n}=5^n-1$ at some stage but 4 doesn't divide LHS Why is $5^n$? __________________________________________ I thinks the solution is: $a_i=2 \Leftrightarrow i=1$ because $a_1 a_2 ... a_n +1$ is odd with every $n > 1$. So $a_1 a_2 ... a_n$ isn't divisible by $4$. On the other hand, $a_1 a_2 ... a_n \neq 4 \Rightarrow$ Q.E.D

$p_1=2$, therefore $p_n,n>1$ are odd and $a_n=p_1....p_n+1=3\mod 4$. Because $p_2=3$ we get $a_n=7\mod 12\to p_n\ge 7$ ($a_n$ had prime divisor form $3\mod 4$ greater than 3. It is at least 7).

$ a_{2}=3$ , then $2$ and $3$ won't appear on the sequence again. For $5$ to appear we need $\prod_{i=1}^{n}a_i\ = 5^{w} - 1$ so $ \prod_{i=1}^{n}a_i\ \equiv 24 (mod\ 100) $ which means $\prod_{i=3}^{n}a_i\ \equiv 4 (mod\ 50)$, but $\prod_{i=3}^{n}a_i\ \equiv 1 (mod\ 2)$ so is odd, so contradiction.

$a_2=3$. Now suppose that for some $n, p_n=5$ then the number $x=p_1p_2...p_{n-1}+1$ could be divisible by 2, 3 or 5. Notice that since $gcd(p_1p_2...p_{n-1},x)=1$ it is impossible for $x$ to have a 2 or 3 in its prime factorization thus; $x= p_1p_2...p_{n-1}+1=5^a$ and therefore $p_1p_2...p_{n-1}=5^a-1=4(5^{a-1}+...+5+1)$. Since all $p_n$ are odd for $n>1$ this equality cannot happen and thus by contradiction, no $p_n$ is equal to $5$

Also Indian RMO 2004 P6. Here