<s>[INST] 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!
 [/INST] First, assume $x$ and $y$ are strictly positive.
$x^4 - y^4 = (x-y)(x+y)(x^2+y^2)$. Of these three factors, only the first can equal 1. So they must be 1, 3, and 5. From 1 and 3 you already get $x=2, y=1$, and then you just have to check that this is a valid solution.
Now we allow the negative solutions, to get $x = \pm 2, y = \pm 1$.
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