3 Variable Diophantine Equation Find all integer solutions to $$x^4 + y^4 + z^3 = 5$$ I don't know how to proceed, since it has a p-adic and real solution for all $p$.
I think that the only one is (2, 2, -3) and the trivial ones that come from this, but I can't confirm it.
 A: After a careful investigation I present some results which might be helpful in the final resolution of this problem. Lets start with the original equation i.e.
$$x^4+y^4+z^3=5$$
It is easy to see that there is no solution of this equation where $x,y$ and $z$ are all positive. Second if $(x,y,z)$ is a solution then so is $(-x,-y,z)$. From this observations we get that $z<0$. Let rewrite the equation in terms of positive values i.e.
$$x^4+y^4-z^3=5$$
where $x,y$ and $z$ are all positive. This is equivalent to 
$$x^4+y^4=5+z^3$$
From this, one gets 
$$x^4+y^4\equiv z^3\mod(5)$$
First we show that $x\equiv 0\mod(5)$ and $y\equiv 0\mod(5)$ iff $z\equiv 0\mod(5)$. 
If $x\equiv 0\mod(5)$ and $y\equiv 0\mod(5)$ then 
$$x^4+y^4\equiv0\mod(5)\Rightarrow 5+z^3\equiv z^3\equiv0\mod(5)\Rightarrow z\equiv0\mod(5)$$ 
Now let $z\equiv0\mod(5)$ then $$5+z^3\equiv0\mod(5)\Rightarrow x^4+y^4\equiv0\mod(5)$$
However for any $x\in\mathbb{Z}$ one has $x^4\equiv0\mod(5)$ or $x^4\equiv1\mod(5)$. Therefore
$$x^4+y^4\equiv0\mod(5)\Leftrightarrow x\equiv y\equiv0\mod(5)$$
 However an inspection of our modified equation yields that we can not have simultaneously $x$ and $y$ divisible by $5$ for otherwise $z\equiv0\mod(5)$ and
$$x^4+y^4\equiv0\mod(5^3)\Rightarrow 5+z^3\equiv0\mod(5^3)$$
which is impossible as $5+z^3\equiv 5\mod(5^3)$.
Knowing that $x$ and $y$ can not be simultaneously divisible by $5$ then $z$ is not divisible by $5$ either. Applying Fermat's little theorem we can rewrite the modified equation as
$$zx^4+zy^4\equiv1\mod(5)$$
 Let say without loss of generality $y\equiv0\mod(5)$ and $x^4\equiv1\mod(5)$ then
$$z\cdot 1+z\cdot 0\equiv1\mod(5)\Rightarrow z\equiv1\mod(5)$$
The other exhaustive case would be $x^4\equiv1\mod(5)$ and $y^4\equiv1\mod(5)$ in which case 
$$z\cdot 1+z\cdot 1\equiv1\mod(5)\Rightarrow 2z\equiv1\mod(5)\Rightarrow z\equiv3\mod(5)$$
In this case a direct inspection for $z=3$ would yield $x=\pm 2$ and $y=\pm 2$.
