All integer solutions for $x^4-y^4=15$ 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?
Thanks!

$x^4-y^4=15\implies (x^2-y^2)(x^2+y^2)=15$. Factorization of $15=1.15$ or $3.5$. Case $1$, $(x^2-y^2)=1\implies (x+y)=1$ and $(x-y)=1$ which have only integer solutions $0,1$ but then $(x^2+y^2)\neq15$, therefore, factorization is $3.5$ which implies $(x^2+y^2)=5$ and $(x^2-y^2)=3$ which gives $x^2=4$ and $y^2=1\implies x=\pm2$ and $y=\pm1$. 
