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## Diophantine $x^p+y^q=(x+y)^r$

Is the equation: $$x^p+y^q=(x+y)^r$$

in integers $x,y,z,p,q,r$ with $p \geq 2,q \geq 2, r \geq 2$ complete solved?

For $(p,q,r)=(n,n,n+1)$ a parametrization is $t=1-s$ and $t(s^n+t^n),s(s^n+t^n),s^n+t^n$

For $(n,n,n-1)$ a rational parametrization is $t=1-s$ and $\frac{t}{s^n+t^n},\frac{s}{s^n+t^n},\frac{1}{s^n+t^n}$

$(3,3,5)$ defines genus $0$ curve with small solutions ${(104, -91, 13),(19005, -18824, 181)}$

Some cases with genus $\leq 1$:

3 3 5 genus= 0 x^4 + 4*x^3*y + 6*x^2*y^2 + 4*x*y^3 + y^4 - x^2 + x*y - y^2
3 5 5 genus= 1 x^4 + 5*x^3*y + 10*x^2*y^2 + 10*x*y^3 + 5*y^4 - x^2
3 3 7 genus= 1 x^6 + 6*x^5*y + 15*x^4*y^2 + 20*x^3*y^3 + 15*x^2*y^4 + 6*x*y^5 + y^6 - x^2 + x*y - y^2
4 2 2 genus= 0 -x^3 + x + 2*y
4 3 4 genus= 0 4*x^3 + 6*x^2*y + 4*x*y^2 + y^3 - y^2
4 4 6 genus= 1 x^6 + 6*x^5*y + 15*x^4*y^2 + 20*x^3*y^3 + 15*x^2*y^4 + 6*x*y^5 + y^6 - x^4 - y^4
5 2 2 genus= 0 -x^4 + x + 2*y
5 5 7 genus= 1 x^6 + 6*x^5*y + 15*x^4*y^2 + 20*x^3*y^3 + 15*x^2*y^4 + 6*x*y^5 + y^6 - x^4 + x^3*y - x^2*y^2 + x*y^3 - y^4
6 2 2 genus= 0 -x^5 + x + 2*y
7 3 3 genus= 1 -x^6 + x^2 + 3*x*y + 3*y^2
7 4 4 genus= 1 -x^6 + x^3 + 4*x^2*y + 6*x*y^2 + 4*y^3
7 2 2 genus= 0 -x^6 + x + 2*y


The $(7,4,4)$ elliptic curve is of rank $1$.

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