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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.

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    $\begingroup$ Under the assumption of the abc conjecture, there are at most finitely many instances where the sum of two coprime fifth powers is powerful. $\endgroup$ – Stefan Kohl Dec 7 '13 at 19:41
  • $\begingroup$ @StefanKohl certainly, I know this. I am asking for a single solution, not for infinitely many. $\endgroup$ – joro Dec 8 '13 at 5:39
  • $\begingroup$ Do you get useful information from looking at x^5 + y^5 = 0 mod p^2? $\endgroup$ – The Masked Avenger Dec 8 '13 at 16:45
  • $\begingroup$ @TheMaskedAvenger I didn't get anything useful mod p^2. $\endgroup$ – joro Dec 9 '13 at 15:25
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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.

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  • $\begingroup$ Thanks, I see. So if a solution would exist, $u,v,s$ are rationals $u'/d,v'/d,s'/d$ and $x,y$ are coprime integers. $\endgroup$ – joro Dec 8 '13 at 8:21
  • $\begingroup$ Thank you. Just noticed that the parametrization is stronger when affine: $x^5+y^5 = z'^{10} t'^6$ $\endgroup$ – joro Dec 8 '13 at 15:29

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