<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] 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 $x=\pm 2, y = \pm 1.$
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