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Prove that the sum of two consecutive odd primes has at least three prime divisors (not necessarily different).

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Well, if $P$ and $Q$ are consecutive odd primes, then $P+Q$ is even, so $2$ divides $P+Q$. Hence, $P+Q=2R$, for some $R$. If $R$ were prime, then $R$ were a prime between $Q$ and $P$, a contradiction.

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