TITLE: Do there exist $29$ consecutive integers so that every of them has exactly $2$ distinct prime factors?
QUESTION [8 upvotes]: Do there exist $29$ consecutive integers, denote $a,a+1,\cdots,a+28$, so that every of them has exactly $2$ distinct prime factors?

For example, $25$ has only one distinct prime factor, and $30$ has $3$ distinct prime factors.
These are my effors:
1.Since they are not divisible by $30$, so $a\equiv 1 \pmod {30}.$
2.I wrote a code (Mathematica 9.0) for this problem:

j = 0; i = 2; While[j < 29 && i < 10^8, 
   If[Length[FactorInteger[i]] == 2, j = j + 1; i = i + 1, j = 0; 
    i = i + 31 - Mod[i, 30]]]; Print[{j, i}]

After runing this program, i find that there are no such numbers when $a<10^8$, it takes about $4$ minutes.
Thanks in advance!

REPLY [9 votes]: If the first number is $30n+1$ we note that:
$$30n+6=6(5n+1)$$$$30n+12=6(5n+2)$$$$30n+18=6(5n+3)$$$$30n+24=6(5n+4)$$
So $5n+r$ is divisible only by $2$ or $3$ for $1 \le r\le4$. $5n+1$ and $5n+2$ are coprime, so one must be a power of $3$ and the other a power of $2$ (neither can be $1$). It is then easy to see that we can't also fit $5n+3$ and $5n+4$ into the same pattern.