# Two Diophantine equations

What is known about solutions in integers of the equations; $x^4 - y^2 = z^6$

I got $x=4st(s^4 - t^4)$ , $z=4st(s^2 - t^2)$ , $y=(4st(s^2 - t^2))^2 (s^4 + t^4 - 6 (st)^2)$

and, the equation $x^2 - y^4 = z^6$

I got $y=4st(s^4 - t^4) , z=4st(s^2 + t^2) , x=(4st(s^2 + t^2))^2 (s^4 + t^4 + 6 (st)^2)$

-
How did you get these solutions? –  Berci Oct 5 '12 at 10:36
@Berci:As I said in another post,I got x=4mn(m+n) , y=4mn(m-n) , z=4mn for x^2 - y^2 = z^3. Then once I put m = (s^2 - t^2)^2 and n=4(st)^2 and got the solution for x^4 - y^2 = z^6 ; and after thinking over some time I put m=(s^2 + t^2)^2 and n=4(st)^2 and got the solution of x^2 - y^4 = z^6 –  Souvik Dey Oct 5 '12 at 10:49

Here is another solution of $x^4 - y^2 = z^6$. Suppose we have a Pythagorean triple $a^2 + b^2 = c^2$, and c is itself a square, say $d^2$. There are many such solutions, a sufficient condition being that c can be factorised into primes of the form $4n+1$, with an even exponent of each prime. Where this is the case each factor can be expressed as a sum of two squares (Fermat's 4n+1 Theorem), then we can use Brahmagupta's identity to express c as a sum of two squares, say $m^2 + n^2$. Then using the standard formula for Pythagorean triples we put a and b equal to $m^2 - n^2$ and $2mn$ (in either order).

Given $a^2 + b^2 = d^4$ we have:

$d^4 - a^2 = b^2$

$(d^4)(b^4) - (a^2)(b^4) = (b^2)(b^4)$

$(bd)^4 - (ab^2)^2 = b^6$

The smallest non-trivial solution of this type has a = 24, b = 7, c = 25, d = 5, yielding $35^4 - 1176^2 = 7^6$. Each such Pythagorean triple yields two solutions by reversing a and b, the second solution in this case with a = 7, b = 24 being $120^4 - 4032^2 = 24^6$. The second solution is an instance of the solution in the question (with s = 2, t = 1), but the first is not (since 4 does not divide 35 or 7). More generally, any solution of the above form with b and d both odd will result in x odd and therefore not be an instance of the solution in the question.

-