Prove that there is some positive integer $n$ such that $n, n +1, n+2,... , n + 200$ are all composite.

Question

Have following approach, which either is inadequate or am unable to amend for this problem:

1. By definition, if $n\le m$, then $n \,|\, m!$.
   
2. Thus for all integers $2\le n\le (m + 1)$, $n \,|\, (m + 1)!$.
   
3. For all integers, $n \,| \,n$ trivially.
   
4. Also, if $n\, |\, a$ and $n\, | \,b\implies n\, | \,(a + b)$, with $a = n, b= (m+1)!$.
   
5. Hence, have : $$2\,|\, 2 +(m+1)!$$ $$3\,|\, 3 +(m+1)!$$ $$\vdots$$ $$n+1\,|\, n +1+(m+1)!$$

Answer 1

What you were doing is correct. What you need is a clever choice for $n$. Choose $n=202!+2$ and it'll work out.

Answer 2

Rephrasing:

$k_1 = 2+(n)!, k_2 =3+(n)!,....$

$...k_{n-1} = (n) +(n)!$.

All of the above numbers $k_i, i=1,2,...,(n-1)$ are composite.

To have $201$ consecutive composite numbers choose

$n= 202$.

This way you can have arbitrarily large gaps between consecutive primes.

Ref:Wiki: Prime gaps

Answer 3

Let $n=(200!)^{3} \in \mathbb{N}$ (absurd number, I know). Then

\begin{alignat}{3} 200 &| n &&= & & (200!)^{3} \\ & &&= & & 200(199!)(200!)^{2} \\ & && & & \\ m &| n+1 &&= & & (200!)^{3}+1 \\ & &&= & & (200!)^{3}-(-1)^{3} \\ & &&= & & [200!-(-1)][(200!)^{2}+(-1)^{1}(200!)^{1}+(-1)^{2}] \\ & &&= & & [200!+1][(200!)^{2}-200!+1] \qquad \qquad \qquad \qquad \qquad \text{ (and so let } m=200!+1\in \mathbb{N}) \\ & && & & \\ 3 &| n+3 &&= & & (200!)^{3}+3 \\ & &&= & & 3[(1 \cdot 2 \cdot 4 \cdot … \cdot 200)(200!)^{2}+1] \\ &... && & & \\ 199 &| n+199 &&= & & (200!)^{3}+199 \\ & &&= & & 199[(1 \cdot 2 \cdot … \cdot 198 \cdot 200)(200!)^{2} + 1] \\ & && & & \\ 200 &| n+200 &&= & & (200!)^{3}+200 \\ & &&= & & 200[(199!)(200!)^{2} + 1]. \\ \end{alignat} QED.