Is it y2 or y2^2?

Justanaccount wrote: Is it y2 or y2^2? In which part !? (Everything that I wrote is the correct form of problem.)

You edited the problem after @2above wrote the message though.

Let's nail this. Suppose $B>A$ and fix $y_1=y_2=4p$, where $p>2$ is a prime. Notice that $x_1^2 + Ay_1^2 = x_2^2 + By_2^2$ implies $(x_1-x_2)(x_1+x_2) = 16p^2(B-A)$. Taking $x_1-x_2=2$ and $x_1+x_2 = 8p^2(B-A)$, we see that \[ (x_1,y_1,x_2,y_2) = \bigl(4p^2(B-A)+1,4p,4p^2(B-A)-1,4p\bigr) \]is a quadruple such that $x_1^2+Ay_1^2$ and $x_2^2+By_2^2$ both realize the same number with $(x_1,y_1)=(x_2,y_2)=1$. As $p$ is arbitrary, the proof is complete.

It has a bash solution, I remembered one of my friends solve it with that bash solution, if you are interested to see that sol, tell me.