Diophantine $x^p+y^q=(x+y)^r$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T13:35:01Z http://mathoverflow.net/feeds/question/77022 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77022/diophantine-xpyqxyr Diophantine $x^p+y^q=(x+y)^r$ joro 2011-10-03T08:23:38Z 2011-10-03T08:23:38Z <p>Is the equation: $$x^p+y^q=(x+y)^r$$</p> <p>in integers $x,y,z,p,q,r$ with $p \geq 2,q \geq 2, r \geq 2$ complete solved?</p> <p>For $(p,q,r)=(n,n,n+1)$ a parametrization is $t=1-s$ and $t(s^n+t^n),s(s^n+t^n),s^n+t^n$</p> <p>For $(n,n,n-1)$ a <strong>rational</strong> parametrization is $t=1-s$ and $\frac{t}{s^n+t^n},\frac{s}{s^n+t^n},\frac{1}{s^n+t^n}$</p> <p>$(3,3,5)$ defines genus $0$ curve with small solutions ${(104, -91, 13),(19005, -18824, 181)}$</p> <p>Some cases with genus $\leq 1$:</p> <pre><code>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 </code></pre> <p>The $(7,4,4)$ elliptic curve is of rank $1$.</p>