Integers as solutions!

$2x^4 -4y^4+8z^4=t^4$

Assume a solution to this is ( $x_0, y_0, z_0, t_0$ ). Therefore, we have:

$2x_0^4 -4y_0^4+8z_0^4=t_0^4$ (1)

$\Leftrightarrow 2(x_0^4 -2y_0^4+4z_0^4)=t_0^4$

We can clearly see that $t^4$ divides 2, so $t_0$ divides 2. Let $t_0=2t_1$ ( $t_1$ is an integer)

$2(x_0^4 -2y_0^4+4z_0^4)=(2t_1)^4=16t_1^4$

$\Leftrightarrow x_0^4 -2y_0^4+4z_0^4=8t_1^4$

It's obvious that $x_0^4$ divides 2, so $x$ divides 2. Let $x_0=2x_1$ ( $x_1$ is an integer)

$(2x_1)^4 -2y_0^4+4z_0^4=8t_1^4 \Leftrightarrow 16x_1^4 -2y_0^4+4z_0^4=8t_1^4$

$\Leftrightarrow 8x_1^4 -y_0^4+2z_0^4=4t_1^4$ (Divide each side by 2)

It's clear that $y_0^4$ divides 2, so $y_0$ divides 2. Let $y_0=2y_1$ ( $y_1$ is an integer).

$8x_1^4 -(2y_1)^4+2z_0^4=4t_1^4 \Leftrightarrow 8x_1^4 -16y_1^4+2z_0^4=4t_1^4$

$\Leftrightarrow 4x_1^4 -8y_1^4+z_0^4=2t_1^4$ (Divide each side by 2)

Again, $z_0^4$ divides 2, so $z_0$ divides 2. Let $z_0=2z_1$ ( $z_1$ is an integer).

$4x_1^4 -8y_1^4+(2z_1)^4=2t_1^4 \Leftrightarrow 4x_1^4 -8y_1^4+16z_1^4=2t_1^4$

$\Leftrightarrow 2x_1^4 -4y_1^4+8z_1^4=t_1^4$ (Divide each side by 2) (2).

Comparing (1) and (2), we can conclude that $(x_1,y_1,z_1,t_1) = (\frac{x_0}{2},\frac{y_0}{2},\frac{z_0}{2},\frac{t_0}{2})$ is another solution.

However, if we keep letting $x_n=2x_{n+1}, y_n=2y_{n+1}, z_n=2z_{n+1}, t_n=2t_{n+1}$ ( $n$ is a natural number), we can always conclude that $(x_{n+1}, y_{n+1},z_{n+1},t_{n+1}) =(\frac{x_n}{2},\frac{y_n}{2},\frac{z_n}{2},\frac{t_n}{2}) = (\frac{x_0}{2^n},\frac{y_0}{2^n},\frac{z_0}{2^n},\frac{t_0}{2^n})$ is another solution for the equation.

We can see that $x_0, y_0, z_0, t_0$ divides $2^n$ for every natural number $n$ . The only solution satisfying this is $(0,0,0,0)$ (since 0 is the only number that divides an infinite series of numbers).

Therefore, the equation has only $\boxed{1}$ solution, that is $(0,0,0,0)$ .