# Find the integers satisfying $a^3+b^3-a^2b^2-(a+b)^2c=0$

Suppose $a,b,c\in \mathbb{Z}_{\ge0}$. How do I find all such $a$, $b$ and $c$ which satisfy $$a^3+b^3-a^2b^2-(a+b)^2 c=0$$

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@Shobhit: More generally, $a=b=0.$ – Charles Jun 28 '13 at 19:36
Please refer to this link: wolframalpha.com/input/… – dreamer Jun 28 '13 at 19:46

If $a>0,b>0,$ $a^3+b^3-a^2b^2-(a+b)^2 c=0$

$\Rightarrow a^3+b^3\equiv a^2b^2 \pmod {(a+b)^2}$

$\Rightarrow ab\equiv 0 \pmod {a+b}$

Denote $\dfrac{ab}{a+b}=k\in \mathbb Z,(a,b)=d,a=a_1 d,b=b_1 d,(a_1,b_1)=1,\dfrac{a_1b_1d}{a_1+b_1}=k.$

Since $(a_1b_1,a_1+b_1)=1,a_1+b_1\mid d.$

Let $d=t(a_1+b_1),$ then $k=a_1b_1t,a=a_1(a_1+b_1)t,b=b_1(a_1+b_1)t.$

$c=\dfrac{a^3+b^3-a^2b^2}{(a+b)^2}=(a_1^2-a_1b_1+b_1^2-ta_1^2b_1^2)t<(a_1^2+b_1^2-a_1^2b_1^2)t=(1-(a_1^2-1)(b_1^2-1))t$

If $a_1>1$ and $b_1>1,$ then $c<0.$ So $a_1=1$ or $b_1=1,$ but no matter in any case, since $\dfrac{c}{t}\in \mathbb Z,$ $\dfrac{c}{t}<1$ means $\dfrac{c}{t}=0.$

Hence if $a>0$ and $b>0$ then $c=0.$

• If $a>0,b>0,c=0,$ then $a^3+b^3-a^2b^2=0,(a_1^3+b_1^3)=da_1^2b_1^2,$ since $(a_1^3+b_1^3,a_1^2b_1^2)=1,$ we have $a_1=b_1=1,d=2.$ We get $a=b=2,c=0.$

• If $a=0,$ then $b^3-b^2c=0, b=0$ or $c=b.$

Hence the only solutions are $(a,b,c)=(0,0,0)(2,2,0)(0,0,n)(0,n,n)(n,0,n).(n>0)$

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