I am wondering if the sum of two non-zero coprime fifth powers can be powerful. There are no small solutions.

Q1 Can the sum of two non-zero coprime fifth powers be powerful?

Got a partial result, possibly wrong.

Consider the surface:

$$ S: x^5+y^5-z^2 t^3=0 $$

According to Magma it is a rational surface with complete parametrization:

```
x,y,z,t=u^8*s^2 + 10*u^7*v*s^2 + 43*u^6*v^2*s^2 + 104*u^5*v^3*s^2 + 155*u^4*v^4*s^2 + 147*u^3*v^5*s^2 + 90*u^2*v^6*s^2 + 36*u*v^7*s^2 + 8*v^8*s^2, u^7*v*s^2 + 9*u^6*v^2*s^2 + 34*u^5*v^3*s^2 + 70*u^4*v^4*s^2 + 85*u^3*v^5*s^2 + 62*u^2*v^6*s^2 + 28*u*v^7*s^2 + 8*v^8*s^2, u^5*s^5 + 10*u^4*v*s^5 + 40*u^3*v^2*s^5 + 80*u^2*v^3*s^5 + 80*u*v^4*s^5 + 32*v^5*s^5, u^10 + 10*u^9*v + 45*u^8*v^2 + 120*u^7*v^3 + 210*u^6*v^4 + 254*u^5*v^5 + 220*u^4*v^6 + 140*u^3*v^7 + 65*u^2*v^8 + 20*u*v^9 + 4*v^10
```

$$ \begin{aligned} x =& (u + v) \cdot s^{2} \cdot (u + 2 v)^{3} \cdot (u^{4} + 3 u^{3} v + 4 u^{2} v^{2} + 2 u v^{3} + v^{4}) \\ y =& v \cdot s^{2} \cdot (u + 2 v)^{3} \cdot (u^{4} + 3 u^{3} v + 4 u^{2} v^{2} + 2 u v^{3} + v^{4}) \\ z =& s^{5} \cdot (u + 2 v)^{5} \\ t =& (u + 2 v)^{2} \cdot (u^{4} + 3 u^{3} v + 4 u^{2} v^{2} + 2 u v^{3} + v^{4})^{2} \end{aligned} $$

Which means $x^5+y^5 = z'^{10} t'^6$.

The inverse map is: $$ u=x^2 t - y^2 t, \; v=x y t + y^2 t, \; s=z t^2 $$

The inverse map implies for integer solution $(x,y,z,t)$, $u,v,s$ must be integers.

From the parametrization $$ \gcd(x,y)=s^{2} \cdot (u + 2 v)^{3} \cdot (u^{4} + 3 u^{3} v + 4 u^{2} v^{2} + 2 u v^{3} + v^{4}) $$

And for coprimality, $|s|=1 ,\; | u + 2 v | =1 ,\; |u^{4} + 3 u^{3} v + 4 u^{2} v^{2} + 2 u v^{3} + v^{4}|=1$ which leads to solution of an univariate polynomial, giving no solution.

Q2 Is the above argument correct?

It would be a proof to FLT for exponent $5$.

The related surface $$ S_7: x^7+y^7-z^3 t^4 = 0 $$

Is rational too.

Q3 Is it true that for a prime $p \ge 5$, the surface $ S_p: x^p + y^p - z^{\lfloor p / 2\rfloor} t^{p - \lfloor p / 2\rfloor} =0 $ is rational?

Magma timeouts. According to a Macaulay2 program even $S$ is not rational.