Diophantine equation in primes

Consider $x_{n+1}=3x_{n}+4y_{n}$ and $y_{n+1}=2x_{n}+3y_{n}$ .

Then $(x_{n+1})^{2}-2(y_{n+1})^{2}=(x_{n})^{2}-2(y_{n})^{2}$ (*)

Setting $x_{1}=3$ and $y_{1}=2$ (which are both solutions to the original equation), it is clear to see that $x_{n+1}$ is always odd and $y_{n+1}$ is always even for all n. Given (*) all values of $(x_{n}, y_{n})$ must also generate solutions to the original equation. Since $y_{n}$ is always even the only prime pair is (3,2) and the answer is 5.

Take mode 4: if x and y are uneven primes the RHS of the Equations is: 1-2×1=-1 but this is contradiction to the LHS which yields to 1 mode 4. therefore we conclude that one of the primes is 2. x cannot be 2 one can see so y=2 and x=3. and x+y=5

Let's assume that $x,y \neq 3$ . Then $1 \equiv 1 - 2 \equiv -1 \mod 3$ . This is a contradiction so at least one of the primes must be equal to 3. If $y = 3$ then $x^2 = 19$ which is a contradiction. If $x=3$ then $-2y^2 = -8 \implies y^2 = 4 \implies y=2$ . This gives us the only solution, $x=3,y=2$ .