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Somedays a ago,when discussed famous problem ($k$ consecutive postive intgers is never a square) with a math teacher,at last,We get the following questions( if we solve following problem,then I think the famous problem have other simple methods to solve it)

Conjecture: for any set $S$ of $n(≥4)$ consecutive positive integers there are two distinct $x,y\in S$ such that pp does not divide $x$ and pp does not divide $y$ where $p(<n)$ is an odd prime?

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    $\begingroup$ Title is badly worded and unhelpful. The post itself needs improvement in grammar and sentence construction. In the current form it is unclear what the hypothesis is and what is to be proved. $\endgroup$ – P Vanchinathan Jan 22 '16 at 5:55
  • $\begingroup$ The conjecture as currently written is not clear. Do you mean: for any set $S$ of $n(\ge4)$ consecutive positive integers there are two distinct $x,y\in S$ such that $p$ does not divide $x$ and $p$ does not divide $y$ where $p(<n)$ is an odd prime? $\endgroup$ – Frentos Jan 23 '16 at 0:35
  • $\begingroup$ @Frentos,Yes,Thanks $\endgroup$ – math110 Jan 23 '16 at 4:10
  • $\begingroup$ Hint. If p divides n, does p divide n-1? does p divide n+1$ does p divide n + 2? $\endgroup$ – fleablood Jan 23 '16 at 5:09
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The conjecture is true.

Let $n \ge 4$ and let $p\le n$ be prime.

Then any sequence of $n$ consecutive integers $a_1,a_2,\cdots,a_n$ will contain a complete set of residues modulo $p$. I.e. for each $r \in { 0,1,\cdots p-1}$ there is some $a \in {a_1,a_2,\cdots,a_n}$ such that $a\equiv r \pmod{p}$.

Case $p=2$: any sequence of $4$ consecutive integers contains exactly $2$ odd numbers. Let $x$ and $y$ be these odd numbers. Then $x \ne y$, $2$ does not divide $x$, $2$ does not divide $y$.

Case $2<p\le n$: Let $x$ be $a_i$ where $a_i \equiv 1 \pmod{p}$. Let $y$ be $a_j$ where $a_j \equiv 2 \pmod{p}$. Then $x \ne y$, $p$ does not divide $x$ and $p$ does not divide $y$.

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We know if $p|m$ and $p|n$ then $p|m \pm n$.

So if $p > 2| m$ the $p$ does not divide $n + 1$ (because $p$ doesn't divide $1 = (n+1) - n$. $p$ does not divide $n - 1$ (same reason) and $p$ does not divide $n + 2$ (same reason; and because $p$ does not divide 2 as $p > 2$) nor $n - 2$ (same reason).

So, let $S = \{m, m + 1, ..... , m + (n-1)\}$. If $p$ divides any of the $m + i$ then $p$ does not divide $m + (i-1), m + (i+1), m + (i+2), m + (i - 2)$.

If $i = 0$ then $m + (i+1), m + (i+2) \in S$.

if $i = n- 1$ then $m + (i -1), m + (i -2) \in S$.

if $0 < i < n -1$ then $m + (i -1), m + (i+1) \in S$.

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