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, $$(k+1)!+2 \\= \{(k+1).k\}.\{(k-1).(k-2)\}.... \\ =2n.n'+2$$, [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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $2$:

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

  • $\begingroup$ ... 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.