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No nontrivial integer solutions to $$ a^5+b^5=c^5+d^5 \qquad (1)$$ are known.

(1) has infinitely many solutions in an extension of $\mathbb{Z}$ (root of $9-15x+37x^2 $ ) resulting from genus 0 curve and the identity: $$ (-2 y - 1)^5 + (x + \frac{1}{3} y - 1)^5- (\frac{1}{3} y - 1)^5 - (x - 2 y - 1)^5 = $$ $$ \left(-\frac{35}{81}\right) \cdot y \cdot (-3 x + 5 y + 6) \cdot x \cdot (9 x^{2} - 15 x y + 37 y^{2} - 18 x + 30 y + 18)$$

I wonder if some similar genus 0 or 1 curve might have points over $\mathbb{Q}$.

Let $p_1,p_2,p_3,p4 \in \mathbb{Q}[x,y]$, $\deg(p_i)=1, p_i \ne -p_j$ and $p_i$ are distinct.

Let $P=p_1^5+p_2^5-p_3^5-p_4^5$.

Q1. Can $P$ have irredicible factor of degree 3?

Q2. Is there a reasons all degree 2 factors of $P$ to not have infinitely many rational points over $\mathbb{Q}$?

Coulnd't solve Q1 by equating coefficients (couldn't solve the system).

Found a lot of genus 0 factors, but all of them didn't have rational points and linear factors gave trivial solutions.

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That genus-zero parametrization gets rediscovered often. It's actually defined over ${\bf Q}$ but alas has no rational points, being birational with the conic $x^2+y^2+z^2=0$. The nice way to see it is to intersect the quintic surface $a^5+b^5+c^5+d^5=0$ with the line $a+b+c+d=0$: the resulting quintic curve decomposes into three lines $a+b=c+d=0$, $a+c=d+b=0$, $a+d=b+c=0$, and a residual conic with $S_4$ symmetry that has no rational or even real roots but does work over ${\bf Q}(i)$ and various other totally imaginary number fields. –  Noam D. Elkies Apr 8 '13 at 16:57
Thank you Elkies. –  joro Apr 9 '13 at 14:42
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