Conjecture on relationship between sum of primes and powers of 2 Let $x$ and $y$ be any odd $\mathbb{N}\geq 2$, and
$n = 2^a$ where $a$ is any $\mathbb{N} \geq 1$:
$$
{x+y \over n} = n^2, n < x < y
$$
when true, then $x$ and $y \in \mathbb{P}$ .

Anyone see any flaw(s) in this, or any obvious explanation which I (most possibly) have overlooked in the heat of the moment?
Is anything known about this; if so, is there an explanation on why this conjecture is true/false?
Edit 1:
Fixed some obvious flaws.
Edit 2:
Fixed conditions.
 A: How about $x=12,y=15,n=3?$
For the update, $x=27,y=189,n=6$
A: Let $X$ and $Y$ be subsets of the naturals. For any given $y\in Y$, the number of pairs $a,b\in X$ such that $a+b=y$ is at most equal to the number of $a\in X$ such that $a\le y$. Thus the number of pairs of primes adding to $m$ grows asymptotically no more than $\frac{\log m}{m}$ by the prime number theorem; at any rate all we need to accept is that primes grow sublinearly, i.e. $o(m)$.
The number of ordered pairs of positive odd numbers that add up to an even number $m$ may be computed exactly as $\frac{m}{2}$, which grows linearly in the value $m$. Whether or not $m$ is restricted to perfect cubes or other special sets is of no meaningful significance. Similarly, if we impose further modular-arithmetic conditions on $a,b$ besides being odd numbers, and force them to always be distinct, and view the pairs as unordered, their count will still grow linearly with $m$ all the same.
It is at face value impossible for a linearly growing collection to fit snugly inside another collection (prime pairs) that grows sublinearly. This is the conceptual explanation of why one would immediately expect a conjecture of this form to be false simply on statistical grounds.
A: $\dfrac{x+y}{n}=n^2 \implies x+y=2^{\alpha+2 \alpha}$
$x+y =2^{3 \alpha} \implies x+y=8^ {\alpha}$
To have $y>x>n$, you need to have $\alpha \ge 2$
I see $x,y \not \in$ prime number set.For instance when $\alpha=2$, $(x,y)=(49,15).(9,55)$ you get only few solutions for $\alpha=2$ because the density of primes to number of possible odds is very high in the close.
What happens when $\alpha >2$? You get more solutions than $\alpha=2$.
