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Let $I_m=[m!+2,m!+m]\cap\mathbb N$ an "interval" of $\mathbb N$; it can obviously be as long as we want and it is easy to prove $I_m$ does not contain any prime. Prove the following: $$\text { if }n^2+(n+1)^2\in I_m\text{ then }4n^2+1\notin I_M$$

Note that if in a large interval $I_m$ could exist $n$ denying what is proposed here, then we would have found a counterexample to the conjecture in here

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  • $\begingroup$ Pardon my ignorance, but why is it "easy to prove $I_m$ does not contain any prime"? If $m$ is very large, the length of the interval is very large at $m-2$, so isn't it likely going to contain a prime? $\endgroup$ – Randall Aug 13 '17 at 1:40
  • $\begingroup$ @Randall $k$ divides $k!+k=k((k-1)!+1)$ $\endgroup$ – uSir470888 Aug 13 '17 at 1:41
  • $\begingroup$ Interesting. I need to think about this. $\endgroup$ – Randall Aug 13 '17 at 1:47
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    $\begingroup$ Wait, no I don't. It's obvious. Thank you. $\endgroup$ – Randall Aug 13 '17 at 1:59
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    $\begingroup$ Hello Randall: everyone is ignorant of many things. Regards. $\endgroup$ – Piquito Aug 13 '17 at 2:00
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Remark(I): At first notice that: $\dfrac{3+\sqrt{9+4}}{2} \leq \dfrac{4+4}{2}=4$ , so for $4 \leq n$ we have:

$$0 \leq n^2-3n+1 \ \ \Longrightarrow \ \ 3n^2+3n+2 \leq 4n^2+1 \ \ \Longrightarrow \ \ \\ \dfrac{3}{2}\Big(2n^2+2n+1 \Big) < 4n^2+1 \ \ \Longrightarrow \ \ \dfrac{3}{2}\Big(n^2+(n+1)^2 \Big) < 4n^2+1 . $$



Remark(II): On the other hand let $4 \leq m$, then we have:

$2 < (m-1)!$, i.e. $1 < \dfrac{(m-1)!}{2}$ multiplying both sides by $m$ we get:

$$m < \dfrac{m!}{2} \ \ \Longrightarrow \ \ m!+m < m!+ \dfrac{m!}{2} = \dfrac{3}{2} m! \ \ \ \ . $$







Suppose on contrary that $4n^2+1 \in I_m $.

Now notice that sicce both of $n^2+(n+1)^2$ and $4n^2+1$

belongs to the interval $[m!+2,m!+m]$, therefor we have:

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ m! < m!+2 \leq n^2+(n+1)^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(III)} \ , \ \ \ \ \ \ \ \ \ \ \ \text {and} \ \ \ \ \\ 4n^2+1 \leq m!+m \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(IV)} .$$


$\color{Red}{\text{First case}}$: Let $m$ and $n$ are both greater or equal than $4$,

i.e. $4 \leq m$ and $4 \leq n$.

In this case we have:

$$\dfrac{3}{2} m! \overset{ \tiny{ \text{III} } }{<} \dfrac{3}{2}\Big(n^2+(n+1)^2 \Big) \overset{ \tiny{ \text{Rmk(I)} } }{<} 4n^2+1 \overset{ \tiny{ \text{IV} } }{\leq} m!+m \overset{ \tiny{ \text{Rmk(II)} } }{<} \dfrac{3}{2} m! \ \ \ \ , $$

so we have: $\dfrac{3}{2} m! < \dfrac{3}{2} m!$ ,which is an obvious contradiction. So this case is immpossible!


$\color{Red}{\text{Second case}}$: Let $4 \leq m$ and $n \leq 3$.

In this case by the (III) inequality we have:

$$26= 4!+2 \leq m! + 2 \overset{ \tiny{ \text{III} } }{\leq} \Big(n^2+(n+1)^2 \Big) \leq 9+16=25 \ , $$

which is again an obvious contradiction!


$\color{Red}{\text{Third case}}$: Let $m \leq 3$ and $4 \leq n$.

In this case by the (IV) inequality we have:

$$265= 4.(4)^2+1 \leq 4n^2+1 \overset{ \tiny{ \text{IV} } }{\leq} m!+m \leq 3!+3=9 \ ,$$

which is again an obvious contradiction!


$\color{Red}{\text{Fourth case}}$: Let $m \leq 3$ and $n \leq 3$.

In this case we have the following sub-cases:

  • $m=3$, then we have: $I_3=[8,9]=\{ 8, 9 \}$. So $n^2+(n+1)^2=8$ or $n^2+(n+1)^2=9$, but none of them have a solution.

  • $m=2$, then we have: $I_2=[6,6]=\{ 6 \}$. So $n^2+(n+1)^2=6$ , but it does'nt have a solution.

  • $m=1$, then we have: $I_1=[3,2]=\phi$.




At the end, it looks, that it was better; if I have been organized the cases as follows:

$\color{Green}{\text{First case}}$: $\color{Yellow}{4 \leq m}$ and $4 \leq n$.


$\color{Green}{\text{Second case}}$: $\color{Yellow}{4 \leq m}$ and $n \leq 3$.


$\color{Purple}{\text{Third case}}$: $\color{Yellow}{m} \color{Orange}{\leq} \color{Yellow}{3}$

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  • $\begingroup$ In the link it has been said that the conjecture has been verified till $10^8$ so the minimum you need for $m$ is $12$ because $11!\lt 10^8$ $\endgroup$ – Piquito Aug 14 '17 at 12:46
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Assume that $2n^2+2n+1 \in [x!+2,x!+x]$ and we want to show that $4n^2+1 \not \in [y!+2,y!+y]$ for some $y>x$.

First see that $4n^2+1 <2(2n^2+2n+1)$ so $\frac{4n^2+1}{2n^2+2n+1} \leq 2$.

Secondly see that $y!+2 > x! +x $ for all $x>2$ , because at least $y=x+1$ so $(x+1) * x! +2 > 2 x! +2x $ for all $x>2$ thus $\frac{y!+2}{x!+x} > 2$.

Which means that $\frac{4n^2+1}{2n^2+2n+1} < 2 < \frac{y!+2}{x!+x}$ in other words if $2n^2+2n+1 \in [x!+2,x!+x]$ then $4n^2+1 \not\in [y!+2,y!+y]$.

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  • $\begingroup$ You have a good idea but some details need clarifying. If $y>x\geq 2$ then $y!+2\geq 2(x!+x)$ because $$y!+2-2(x!+x)\geq x!(x+1)+2-x!\cdot 2-2x=(x-1)(x!-2)\geq 0.$$..... So if $y>x\geq 2$ and $n^2+(n+1)^2\in I_x$ then $$4n^2+1<2 (n^2+(n+1)^2)\leq 2(x!+x)\leq y!+2.$$ (Note the "$<$" in that.) $\endgroup$ – DanielWainfleet Aug 14 '17 at 16:18
  • $\begingroup$ @DanielWainfleet thank you, this was what in mind but i really have a hard time explaining in English since it's not my native-language, non the less its a complete proof. $\endgroup$ – Ahmad Aug 14 '17 at 20:10
  • $\begingroup$ It occurred to me that modern mathematical "grammar" is very much like English grammar. In particular, the order of words is crucial. And that this is not an objective necessity but a method of communicating. In Latin there are some (short) sentences that are unchanged by any permutation of the words, because of its "case-endings". In German two or more words can be joined to make one word, which can remove ambiguities that occur in English. In English there is a grammatically correct sentence in which the last 6 words, in order, are "to from out of up for". $\endgroup$ – DanielWainfleet Aug 15 '17 at 17:12

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