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!
 A: $$(x^2-y^2)+(x^2+y^2)=2x^2=3+5=8\Rightarrow x^2=4\Rightarrow x=2,-2$$
$$(x^2+y^2)-(x^2-y^2)=2y^2=5-3=2\Rightarrow y^2=1\Rightarrow y=1,-1$$
Substitute the values to check that these are indeed the soloutions.
A: 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 $x = \pm 2, y = \pm 1$.
A: Consider $f(x) = (x+1)^4 - x^4 = 4x^3 + 6x^2 + 4x + 1$. Then $f'(x) = 12x^2 + 12x + 4 = 3(2x+1)^2 + 1$, which is always positive. So f(x) is always increasing, meaning that for y more than 1 we don't have any valid solutions. Similarly you can go into the negatives and see that y cannot be less than -1. Now you only have to check a very limited set of values, for y going from -1 to 1 and x going from -2 to 2. It is easy to see now that the only solutions are $x=\pm 2, y = \pm 1.$
A: $x^4-y^4=15\implies (x^2-y^2)(x^2+y^2)=15$. 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 $y=\pm1$. 
