A question related to the existence of a prime 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
 A: 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 $n \leq 3$. 

$\color{Purple}{\text{Third case}}$: 
$\color{Yellow}{m} \color{Orange}{\leq} \color{Yellow}{3}$
A: 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 $4n^2+1 \not\in [y!+2,y!+y]$.
