I'm trying to find all the integer solutions for $x^4-y^4=15$. I know that the options are $x^2-y^2=5, x^2+y^2=3$, or $x^2-y^2=1, x^2+y^2=15$, or $x^2-y^2=15, x^2+y^2=1$, and the last one $x^2-y^2=3, x^2+y^2=5$.
Only the last one is valid. $x^2+y^2=15$ is not solvable since the primes which have residue $3$ modulo $4$ is not of an equal power.
One particular solution for $x^2-y^2=3, x^2+y^2=5$, is $x_0=2, y_o=1$. How do I get to an expression of a general solution for this system?