1987 OIM Problems/Problem 4

Problem

We define the sequence $p_n$ as follows: $p_1=2$ , and for all $n\ge2$ , $p_n$ is the greatest prime divisor of the expression: $p_1p_2p_3\cdots p_{n-1} +1$ Prove that $p_n\ne5$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

~additional modifications by eevee9406

Solution

We get $p_2=3$ from the definition; thus for all $k\ge 3$ , we must have $2,3\nmid p_k$ . But if we assume for the sake of contradiction that there exists $n$ such that $p_n=5$ , then $p_1p_2\cdots p_{n-1}=5^m-1$ for some positive integer $m$ because $5$ is the minimal prime divisor of $p_1p_2\cdots p_{n-1}$ since $2$ and $3$ are not prime divisors, but $5$ must be the maximal prime divisor by definition. However, taking mod $4$ reveals that the LHS is divisible by $2$ but not $4$ whereas the RHS is divisible by $4$ , a contradiction.

~ eevee9406

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1987 OIM Problems/Problem 4

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