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.
2026-03-30 17:39:35.1774892375
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3 Variable Diophantine Equation
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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 $y=\pm 2$.
Solutions to $$ z^3 + 5 = k^4 + s^2 $$ are reasonably frequent