For $x+y=n$, if $V'-V=3$ then $\large \frac{n}2$ is prime.

Define $V$ and $V'$ as the number of visible and non-visible lattice points from the origin respectively.

Conjecture

For $x+y=n, (x,y)\in\mathbb{Z}^2$, if $V'-V=3$ then $\large \frac{n}2$ is prime.

Examples

Let $n=6,V=2\textrm{ (green)},V'=5 \textrm{ (purple)}$ then $3$ is prime.

Similarly for

n=10,V=4,V'=7 then 5 is prime.

n=14,V=6,V'=9 then 7 is prime.

n=22,V=10,V'=13 then 11 is prime.

n=26,V=12,V'=15 then 13 is prime.

n=34,V=16,V'=19 then 17 is prime.

n=38,V=18,V'=21 then 19 is prime.

...

A point $(x,y)$ is visible if and only if $\gcd(x,y)=1$. If $x+y=n$, then $\gcd(x,y)=\gcd(x,x+y)=\gcd(x,n)$. So, $V$ is the number of $0\leq x\leq n$ with $\gcd(x,n)=1$, and $V'=(n+1-V)$. So, you're asking for when $$3=V'-V\Longleftrightarrow V'=\frac n2+2.$$ If this is true, then $n$ must be even, so let $n=2k$ for some $k$. Then, if $x$ is even, $\gcd(x,n)\geq 2$, so $(x,n-x)$ cannot be visible. This accounts for $k+1$ points $(0,2k),(2,2k-2),\dots,(2k-2,2),(2k,0)$. If $k$ is a power of $2$, then these are the only non-visible points, so $V'=k+1$, not $k+2$. Otherwise, if $k$ is not prime, there is some odd prime $p$ which divides $k$; in this case, $(p,n-p)$ and $(n-p,p)$ are distinct non-visible points that we haven't already counted. This gives that $V'\geq k+3$, so $V'\neq k+2$. So, $k$ must be prime, as desired. (The last non-visible point is $(k,k)$, and one can show there aren't any others, which gives the converse of the statement for whenever $n/2$ is an odd prime.)

Note: for $n>1$, $V$ is the same as the Euler totient function $\varphi(n)$ of $n$. Your conjecture can be rephrased as requiring $\varphi(n)=n/2-1$ and asking if $n$ is twice a prime.