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Prove that there are n\, consecutive composite numbers, for any n>0\,.

Let m=n+1\,. Consider the sequence of n\, numbers

m!+2,m!+3,\ldots ,m!+m\,

Then for any k\, such that 2\leq k\leq m\,

m!+k\, =\left(1\cdot 2\cdot \ldots \cdot (k-1)\cdot k\cdot (k+1)\cdot \ldots \cdot m\right)+k\,
=k\left[\left(1\cdot 2\cdot \ldots \cdot (k-1)\cdot (k+1)\cdot \ldots \cdot m\right)+1\right]\,

Therefore, m!+k\, is composite, and since k\, was arbitrary, every number in the sequence is composite.

As an interesting side note, now consider taking n\to \infty \,.

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