Positive Integer Cover Who knows whether or not union of the following 5 equations for $x,y=0,1,2,3,4,...$ covers natural numbers fully after a certain integer?
$A=10xy+x+y$ (non zero $x$ and $y$)
$B=10xy+7x+3y+2$
$C=10xy+9x+9y+8$
$D=10xy+x+3y$ (non zero $y$)
$E=10xy+7x+9y+6$
(Tips: we know the set $A\cup B\cup C$ will never cover $\Bbb N$ fully, whether one by one correspondence of the elements of the set $D\cup E$ will help it to cover $\Bbb N$ fully after a certain integer?)
Let $A,B,C,D,E$ be given by
\begin{align*}             
A&amp;=\{10xy+x+y{\,\mid\,}x,y\in\mathbb{N}\}\\[2pt]      
B&amp;=\{10xy+7x+3y+2{\,\mid\,}x,y\in\mathbb{N}_0\}\\[2pt]          
C&amp;=\{10xy+9x+9y+8{\,\mid\,}x,y\in\mathbb{N}_0\}\\[2pt]          
D&amp;=\{10xy+x+3y{\,\mid\,}x\in\mathbb{N}_0,y\in\mathbb{N}\}\\[2pt]         
E&amp;=\{10xy+7x+9y+6{\,\mid\,}x,y\in\mathbb{N}_0\}\\[2pt]
\end{align*}
where $\mathbb{N}$ denotes the set of positive integers, and $\mathbb{N_0}$ denotes the set of nonnegative integers.

Claim:$\;$The following statements are equivalent:

$A\cup B\cup C\cup D\cup E$ contains all but finitely many positive integers.$\\[4pt]$
There are only finitely many positive integers $n$ such that $10n+1$ and $10n+3$ are both prime.

Proof:

Let $n\in A\cup B\cup C\cup D\cup E$.

Consider cases . . .
\begin{align*}
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
&amp;n\in A\\[2pt]
\iff\;&amp;10xy+x+y=n,\;x,y\in\mathbb{N}\\[2pt]
\iff\;&amp;(10x+1)(10y+1)=10n+1,\;x,y\in\mathbb{N}\\[0pt]
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
&amp;n\in B\\[2pt]
\iff\;&amp;10xy+7x+3y+2=n,\;x,y\in\mathbb{N}_0\\[2pt]
\iff\;&amp;(10x+3)(10y+7)=10n+1,\;x,y\in\mathbb{N}_0\\[0pt]
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
&amp;n\in C\\[2pt]
\iff\;&amp;10xy+9x+9y+8=n,\;x,y\in\mathbb{N}_0\\[2pt] 
\iff\;&amp;(10x+9)(10y+9)=10n+1,\;x,y\in\mathbb{N}_0\\[0pt]
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
&amp;n\in D\\[2pt]
\iff\;&amp;10xy+x+3y=n,\;x\in\mathbb{N}_0,y\in\mathbb{N}\\[2pt]
\iff\;&amp;(10x+3)(10y+1)=10n+3,\;x\in\mathbb{N}_0,y\in\mathbb{N}\\[0pt] 
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
&amp;n\in E\\[2pt]
\iff\;&amp;10xy+7x+9y+6=n,\;x,y\in\mathbb{N}_0\\[2pt]
\iff\;&amp;(10x+9)(10y+7)=10n+3,\;x,y\in\mathbb{N}_0\\[0pt] 
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
\end{align*}
It follows that $n\in A\cup B\cup C\cup D\cup E$ if and only if at least one of $10n+1,10n+3$ factors non-trivially, which proves the claim.

But given the current state of the Twin Prime conjecture, there may or may not be infinitely many positive integers $n$ such that $10n+1$ and $10n+3$ are both prime, but whichever is the case, a proof is likely not yet achievable.
Let $A,B,C,D,E$ be given by
\begin{align*}             
A&amp;=\{10xy+x+y{\,\mid\,}x,y\in\mathbb{N}\}\\[2pt]      
B&amp;=\{10xy+7x+3y+2{\,\mid\,}x,y\in\mathbb{N}_0\}\\[2pt]          
C&amp;=\{10xy+9x+9y+8{\,\mid\,}x,y\in\mathbb{N}_0\}\\[2pt]          
D&amp;=\{10xy+x+3y{\,\mid\,}x\in\mathbb{N}_0,y\in\mathbb{N}\}\\[2pt]         
E&amp;=\{10xy+7x+9y+6{\,\mid\,}x,y\in\mathbb{N}_0\}\\[2pt]
\end{align*}
where $\mathbb{N}$ denotes the set of positive integers, and $\mathbb{N_0}$ denotes the set of nonnegative integers.

Claim:$\;$The following statements are equivalent:

$A\cup B\cup C\cup D\cup E$ contains all but finitely many positive integers.$\\[4pt]$
There are only finitely many positive integers $n$ such that $10n+1$ and $10n+3$ are both prime.

Proof:

Let $n\in A\cup B\cup C\cup D\cup E$.

Consider cases . . .
\begin{align*}
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
&amp;n\in A\\[2pt]
\iff\;&amp;10xy+x+y=n,\;x,y\in\mathbb{N}\\[2pt]
\iff\;&amp;(10x+1)(10y+1)=10n+1,\;x,y\in\mathbb{N}\\[0pt]
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
&amp;n\in B\\[2pt]
\iff\;&amp;10xy+7x+3y+2=n,\;x,y\in\mathbb{N}_0\\[2pt]
\iff\;&amp;(10x+3)(10y+7)=10n+1,\;x,y\in\mathbb{N}_0\\[0pt]
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
&amp;n\in C\\[2pt]
\iff\;&amp;10xy+9x+9y+8=n,\;x,y\in\mathbb{N}_0\\[2pt] 
\iff\;&amp;(10x+9)(10y+9)=10n+1,\;x,y\in\mathbb{N}_0\\[0pt]
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
&amp;n\in D\\[2pt]
\iff\;&amp;10xy+x+3y=n,\;x\in\mathbb{N}_0,y\in\mathbb{N}\\[2pt]
\iff\;&amp;(10x+3)(10y+1)=10n+3,\;x\in\mathbb{N}_0,y\in\mathbb{N}\\[0pt] 
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
&amp;n\in E\\[2pt]
\iff\;&amp;10xy+7x+9y+6=n,\;x,y\in\mathbb{N}_0\\[2pt]
\iff\;&amp;(10x+9)(10y+7)=10n+3,\;x,y\in\mathbb{N}_0\\[0pt] 
&amp;\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{--------------------------------------------------------------------------------}\\[0pt]
\end{align*}
It follows that $n\in A\cup B\cup C\cup D\cup E$ if and only if at least one of $10n+1,10n+3$ factors non-trivially, which proves the claim.

But given the current state of the Twin Prime conjecture, there may or may not be infinitely many positive integers $n$ such that $10n+1$ and $10n+3$ are both prime, but whichever is the case, a proof is likely not yet achievable.