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An exam paper has given this question.

Let $k$ be a positive integer. We have to prove that $$(k+1)!+2, (k+1)!+3,...,(k+1)!+k, (k+1)!+(k+1)$$ are $k$ consecutive composite intergers.

All I need a proof verification.

Let $k>0$. Let us assume that, $$(k+1)!+2, (k+1)!+3,...,(k+1)!+k, (k+1)!+(k+1)$$ are $k$ consecutive prime numbers. Hence $k>1$.

Hence from $(k+1)!+2$ we get, $$(k+1)!+2 \\= \{(k+1).k\}.\{(k-1).(k-2)\}....4.3.2.1+2 \\ =2n.n'+2$$, [where $n=(k+1)k, n'=(k-1)!, M=n.n'+1$] $$2(n.n'+1)=2.M$$

Now, $M$ divides $(n.n'+1)$, $M$ divides $(k+1)!+2$, hence $(n.n'+1)$ divides $(k+1)!+2$. Hence, $(k+1)!+2$ is not a prime, it's composite.

Similarly, $(k+1)!+3$, $(k+1)!+4$ are composite. Let, $(k+1)!+k$ be composite but assume $(k+1)$ and $(k+1)!+(k+1)$ are prime.

By Wilson's theorem, $$(k+1-1)!\equiv -1\mod {k+1} \\ \Rightarrow k!\equiv k \mod {k+1} \\ \Rightarrow k(k+1)(k-1)!\equiv 0 \mod {k+1} \\ \Rightarrow (k+1)!+(k+1)\equiv (k+1)\equiv 0 \mod {k+1}$$

$(k+1)$ divides $(k+1)$ and $(k+1)!+(k+1)$ hence $(k+1)$ is composite, hence $(k+1)!+(k+1)$. Hence, $$(k+1)!+2, (k+1)!+3,...,(k+1)!+k, (k+1)!+(k+1)$$ are $k$ consecutive composite intergers.

Does it correctly complete the proof? Any suggestion or help is highly appreciated.

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    $\begingroup$ You don't need Wilson $\endgroup$ – AgentS Mar 9 '18 at 14:19
  • $\begingroup$ Why do you assume at the start that they are all prime numbers? You don't really use that assumption, and if you found a contradiction it wouldn't show that the set is all composite in any case. $\endgroup$ – Joffan Mar 9 '18 at 14:23
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    $\begingroup$ You are overthinking this... The key fact of your assessment of $(k+1)!+2$ should be that it is divisible by $2$; and that $(k+1)!+3$ is divisible by $3$ etc. $\endgroup$ – Joffan Mar 9 '18 at 14:25
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    $\begingroup$ Reminds me of myself when I first learned proof by contradiction, I used to attempt all proofs by using contradiction and piss everyone ;) $\endgroup$ – AgentS Mar 9 '18 at 14:28
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Your proof is too complicated.

The numbers are $(k+1)!+t$ with $2 \le t \le k+1$ and so $(k+1)!+t$ is clearly a multiple of $t \ge 2$, so not prime because the quotient is at least $2$:

$$ t \le k+1 \le (k+1)! \implies 2t \le (k+1)!+t \implies 2 \le \frac{(k+1)!+t}{t} $$

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  • $\begingroup$ ... or, $t$ divides $a_t:=(k+1)!+t$ and also $t$ divides $(k+1)!<a_t$ therefore $a_t$ is composite. $\endgroup$ – Joffan Mar 9 '18 at 14:31
  • $\begingroup$ @lhf, Just to clear my confusion I put this question before you . . $t \le k+1 \le (k+1)! \implies 3t \le (k+1)!+2t \implies 3 \le \frac{(k+1)!+2t}{t}$ ... how would you treat the $2t$ there? $\endgroup$ – vbm Mar 9 '18 at 14:45
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    $\begingroup$ @thevbm, add $t$ to both sides of $t \le (k+1)! $. $\endgroup$ – lhf Mar 9 '18 at 14:48
  • $\begingroup$ Really helpful.. thank you $\endgroup$ – vbm Mar 9 '18 at 14:49

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