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.



Can the sum of two non-zero coprime fifth powers be powerful?
Your formulas are only inverse up to rescaling. For example, $(u,v,s) = (1,1,1)$ maps to $(x,y,z,t) = (594, 297, 243, 1089)$ maps to $(u,v,s) = (288178803, 288178803, 288178803)$. More generally, 
$$ (u,v,s) \mapsto (x,y,z,t) \mapsto (\Delta u,\Delta v,\Delta s)$$
where 
$$ \Delta = s^4 (u + 2 v)^9 (u^4 + 3 u^3 v + 4 u^2 v^2 + 2 u v^3 + v^4)^4. $$


So $(x,y,z,t)$ integral doesn't imply $(u,v,s)$ integral, it only implies that $(\Delta u,\Delta v,\Delta s)$ is integral.




---


Yes, $x^p+y^p = z^{\lfloor p/2 \rfloor} t^{\lceil p/2 \rceil}$ is rational. Put $p=2m+1$. I assume, since you call it a surface, you are considering this as a projective equation. Dehomogenize with respect to $t$ to get $x^{2m+1}+y^{2m+1} = z^m$.


In the open chart where $xyz \neq 0$, we can write this as $(x/y)^{2m+1} + 1 = z^m/y^{2m+1}$. Make the change of coordinates


$$a=xy^{-1},\ b=y^{-(2m+1)} z^m,\ c= y^2 z^{-1}$$
with inverse
$$x=ab^{-1} c^{-m},\ y=b^{-1} c^{-m},\ z=b^{-2} c^{-(2m+1)}.$$


In the new coordinates, we have $a^{2m+1}+1 = b$, with the obvious parametrization $(a,b,c) = (s,s^{2m+1}+1, t)$. 


This is a standard trick for finding good rational forms of high degree equations with few variables: Find the lattice generated by the differences of the exponents -- in this case, the $\mathbb{Z}$-span of $(2m+1,-(2m+1),0)$ and $(0,-(2m+1), m)$. Saturate it -- we get the $\mathbb{Z}$-span of $(1,-1,0)$ and $(0,-(2m+1),m)$ and complete to a basis $\begin{pmatrix} 1 & -1 & 0 \\ 0 & - (2m+1) & m \\ 0 & 2 & -1 \end{pmatrix}$ of $\mathbb{Z}^3$. Inverting this determinant $1$ matrix gave me the formulas for $(x,y,z)$ in terms of $(a,b,c)$, and the equation become much lower degree in the new variables.


