given a positive integer $n\geq 2$, we have a positive integer $m$ such that $m+2,m+3,\dots m+n$ are composite. (TIFR exam $2012$)

Question

Question is to prove that :

given a positive integer $n\geq 2$, we have a positive integer $m$ such that $m+2,m+3,\dots m+n$ are composite.

I tried checking for small numbers to see if there is any pattern...

for $n=3$, i have $m=6$ with $m+2=8,m+3=9$ are composite

for $n=4$, i have $m=12$ with $m+2=14,m+3=15,m+4=16$ are composite

for $n=5$, i have $m=23$ with $m+2=25,m+3=26,m+4=27,m+5=28$ are composite

I do not see any pattern which assure me existence of $m$ for a general $n$

I would be thankful if some one can help me out with some hint.. Thank you

Answer 1

For $m=n!$ we have:

      $n!+2$ has $2$ as factor. So, it is composite.

      $n!+3$ has $3$ as factor. So, it is composite.

In general,

      $n!+k$ has $k$ as factor (for $k \leq n$). So, it is composite.

This means $n!+2,n!+3,\dots,n!+n$ are composite.

Thus, given a positive integer $n\geq2$, we have a positive integer $m$ such that $m+2,m+3,\dots,m+n$ are composite.