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Is there a way to pick a $k$ such that $p_0p_1 + ik$ is always a product of two primes? where $p_0, p_1$ are primes, $k < pq$, and $i = 0,1,2,3....$. In other words, you pick $k$ such that no matter what $i$ I use, $p_0p_1 + ik$ is still a product of two primes.

I tried with many examples, and saw that it always failed but I can't prove it logically. Would anyone give me a hint? Thanks in advance.

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$p_0p_1+(p_0p_1)k = p_0p_1 (1 + k)$, no? – Jon Aug 10 '11 at 1:26
@Jon: Is $(1+k)$ prime? We don't know, plus $p_0p_1$ are already two primes. – Chan Aug 10 '11 at 1:28
My point is that if $k \neq 0$, then $p_0p_1 + ik$ always has at least 3 prime factors for $i = p_0p_1$. – Jon Aug 10 '11 at 1:31
@Jon: Ah, I see. Thanks a lot. – Chan Aug 10 '11 at 1:51
@Jon: Why don't you make that an answer? – JavaMan Aug 10 '11 at 1:57
up vote 4 down vote accepted

If $k$ is non-zero, then $p_0p_1 + ik$ will have at least three prime factors when $i$ is a non-zero multiple of $p_0p_1$, since in that case $p_0p_1 + ik = p_0p_1 + np_0p_1k = p_0p_1(1+nk)$.

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HINT $\ $ The arithmetic progression $\rm\ p_0\:p_1 + k\ \mathbb Z\ $ contains elements that are divisible by any integer $\rm\:n\:$ coprime to $\rm\:k\:.\:$ Namely $\rm\: n\ |\ p_0\:p_1 + k\ x\ $ for $\rm\ x\equiv -p_0\:p_1/k\pmod{n}\:.\:$ In particular it has elements divisible by arbitrarily many prime factors via $\rm\:n =\:$ products of primes coprime to $\rm\:k\:.$

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