Are there infinitely many $n$ where the (number of non coprime pairs) $-$ (number of coprime pairs) is 3? For even $n \in \mathbb{N}$, suppose we have this sequence of pairs
$(0,n),(1,n-1),(2,n-2),\dots,(\frac{n}{2},\frac{n}{2}),\dots,(n-2,2),(n-1,1),(n,0)$
Define the Difference formula as
$D =$ (number of non coprime pairs) $-$ (number of coprime pairs)
Question
Are there infinitely many $n$ such that $D=3$?
Examples
Let $n=6$, this sequence has $D=5-2=3$
$(0,6),(1,5),(2,4),(3,3),(4,2),(5,1),(6,0)\\$
Let $n=10$, this sequence has $D=7-4=3$
$(0,10),(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1),(10,0)\\$
Might be related to the Euler totient function?
 A: By Bezout's lemma, two integers $x$ and $y$ are coprime if and only if there exist some integers $a$ and $b$ such that $ax+by=1$. So $k$ and $n-k$ are coprime if and only if $ak+b(n-k)=1$ or equivalently if $(a-b)k+bn=1$ so if and only if $n$ and $k$ are coprime.
The number of $k$ coprime with $n$ is indeed the Euler totient function $\phi(n)=|\{1\leq \ell\leq n-1: gcd(n,\ell)=1\}|$. Note that the pairs $(0,n)$ and $(n,0)$ are never coprime with your setup, so what you are asking can be rephrased as follows: Are there infinitely many $n$ such that $n+1-2\phi(n)=3$ or equivalently such that $n=2+2\phi(n)$. This is because the number of non-coprime pairs is just $n+1-\phi(n)$.
Now the equation $n=2+2\phi(n)$ immediately tells us that $n$ must be even, say $n=2m$. But then
\begin{equation}
\phi(2m)=m-1. (*)
\end{equation}
Since $n$ is even, it will definitely not be coprime with any of the numbers $\{2,4,\dots,2m\}$, which leaves at most $m$ numbers it could be coprime with, namely $\{1,3,\dots,2m-1\}$. But then $(*)$ implies that out of all of these exactly one number should be not coprime with $n=2m$ and all others should be coprime. This already tells you that $m$ should be prime.
Indeed, if we assume $n=2p$ with $p$ an odd prime, we can use the Euler totient product rule (you can see the proof on wiki) to explicitly compute
\begin{equation}
\begin{split}
\phi(2p) & = 2p \prod_{q|(2p), q \text{ prime}} \big(1-\frac{1}{q}\big)\\
& = 2p \big(1-\frac{1}{2}\big)\big(1-\frac{1}{p}\big)\\
& = 2p \cdot \frac{1}{2}\cdot \frac{p-1}{p}\\
& = p-1. 
\end{split}
\end{equation}
Therefore $(*)$ is satisfied whenever $n=2p$ with $p$ an odd prime number. So clearly there are infinitely many solutions.
A bit more can be said. If $n$ is not of the form $n=2p$ for some odd prime $p$, then we have two cases to look at (keeping in mind that $n$ must be even to satisfy $(*)$).
Case 1
$n=2^j a b$ for some power $j\geq 1$ and some odd numbers $a,b\geq 3$. In this case, it is trivial to see that both both $a,ab\in\{1,3,\dots,n-1\}$ and neither of them is coprime with $n$, so immediately $\phi(n)\leq \frac{n}{2}-2$, so $(*)$ is violated.
Case 2
$n=2^j p$ for some odd prime $p$ and some power $j\geq 2$. In this case, we can again explicitly compute $\phi(n)=2^{j-1} (p-1)$ using the same product formula. However, according to $(*)$, $\phi(2^j p)$ should be $2^{j-1}p -1$. Since $j\geq 2$, these two expressions do not match. The discrepancy comes from the fact that some odd multiples of $p$ will also belong to the set $\{1,3,\dots,n-1\}$. For example, if $j=2$, then $3p$ is not coprime with $2^2 p$.
These two cases imply that $(*)$ is satisfied if and only if $n=2p$ for some odd prime $p$.
